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[SOLVED] C - assigment makes integer from pointer without a cast warning
Thread: [solved] c - assigment makes integer from pointer without a cast warning, thread tools.
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I know it's a common error, and I've tried googling it, looked at a bunch of answers, but I still don't really get what to do in this situation.... Here's the relevant code: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char path[51]; const char* home = getenv( "HOME" ); strcpy( path, argv[1] ); path[1] = home; return 0; } -- there is more code in the blank lines, but the issue's not there (I'm fairly sure), so didn't see the point in writing out 100 odd lines of code. I've tried some stuff like trying to make a pointer to path[1], and make that = home, but haven't managed to make that work (although maybe that's just me doing it wrong as opposed to wrong idea?) Thanks in advance for any help
Re: C - assigment makes integer from pointer without a cast warning
path[1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char. You get the warning because you try to assign a char* to a char. Note also the potential to overflow your buffer if the content of the argv[1] argument is very long. It's usually better to use strncpy.
Last edited by r-senior; March 10th, 2013 at 03:03 PM . Reason: argv[1] would overflow, not HOME. Corrected to avoid confusion.
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You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that.
Last edited by Christmas; March 10th, 2013 at 09:51 PM .
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Originally Posted by Christmas You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that. Excellent point. I've basically fixed my problem by reading up on pointers again (haven't done any C for a little while, so forgot some stuff), and doing: Code: path[1] = *home; the code doesn't moan at me when I compile it, and it runs okay (for paths which aren't close to 51 at least), but after reading what you read, I just wrote a quick program and found out that getenv("HOME") is 10 characters long, not 1 like I seem to have assumed, so I'll modify my code to fix that.
Yes, getenv will return the path to your home dir, for example /home/user, but path[1] = *home will still assign the first character of home to path[1] (which would be '/').
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Makes Pointer From Integer Without a Cast: Fix It Now!
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Our research team has ensured that this will be your best resource on these error messages, and you won’t have to read another article about it again. With that said, launch your code that’s showing the error messages, and let’s fix it for you.
JUMP TO TOPIC
– You Assigned an Integer to a Pointer
– you want to convert an integer to a pointer, – you passed a variable name to the “printf()” function, – you used “struct _io_file” in a wrong way, – you copied a string to an invalid location, – you’re setting a pointer to a different type, – use equal data types during assignment, – ensure the pointer and integer have the same sizes, – pass a “format string” to the “printf()” function, – use a function that returns a pointer to “struct _io_file”, – copy the string to character array or character pointer, – assign pointers of compatible types, why your code made a pointer from an integer without a cast.
Your code made a pointer from an integer without a cast because you assigned an integer to a pointer or you want to convert an integer to a pointer. Other causes include the passing of a variable name to the “printf()” function and using “struct _io_file in the wrong way.
Finally, the following are also possible causes:
- You copied a string to an invalid location
- You’re setting a pointer to a different type
If you assign an integer to a pointer, it will lead to the “ assignment makes pointer from integer without a cast c programming ” error. For example, in the following, the “num_array” variable is an unsigned integer type while “tmp_array” is a pointer.
Later, an error will occur when the “for” loop tries to copy the values of the “num_array” to the “tmp_array” pointer using a variable assignment.
For example, in the following, the “theta” variable is an integer while “ptr_theta” is a pointer. Both have different sizes, and you can’t convert the integer to a pointer.
If you pass a variable name to the “printf()” function, that’s when the compiler will throw the “ passing argument 1 of ‘printf’ makes pointer from integer without a cast ” error.
For example, in the following, “num_val” is an integer variable, and the code is trying to print it using the “printf()” function.
When you assign an integer to a pointer of “struct _io_file”, that’s when you’ll get the “ assignment to file aka struct _io_file from int makes pointer from integer without a cast ” error. For example, in the following code, “FILE” is a “typedef” for “struct _io_file”. This makes it a pointer to “struct _io_file”.
Later, the code assigned it to the integer “alpha,” and this will lead to an error stated below:
Now, in the following, the code is trying to copy a string (“source”) into an integer (“destination”), and this leads to an error because it’s not a valid operation:
When your code sets a pointer to a different type, your compiler will show the “ incompatible pointer type ” error. In the following code, “pacifier” is an integer pointer, while “qwerty” is a character pointer.
Both are incompatible, and your C compiler will not allow this or anything similar in your code.
How To Stop a Pointer Creation From an Integer Without a Cast
You can stop a pointer creation from an integer without a cast if you use equal data types during the assignment or ensure the integer and pointer have the same sizes. What’s more, you can pass a “format string” to “printf()” and use a function that returns a pointer to “struct _io_file”.
- Copy the string to a character array or character pointer
- Assign pointers of compatible types
During a variable assignment, use variables of equal data types. In our first example, we assigned a pointer (uint8_t *tmp_array) to an unsigned integer (uint8_t num_array) and it led to a compile-time error.
Now, the following is the revised code, and we’ve changed “tmp_array” to a real array. This will prevent the “ gcc warning: assignment makes pointer from integer without a cast ” error during compilation.
Now, the fix is to use “intptr_t” from the “stdint.h” header file because it guarantees that the pointer and integer will have the same sizes. We’ve used it in the following code, and you can compile it without an error.
To fix the “pointer to integer without a cast” in the “printf()” function, pass a “format string” as its first argument. How you write the “format string” depends on the variable that you’ll print. In the following updated code, “num_val” is an integer, so the “format string” is “%d”.
When you’re using “FILE” from the “stdlib.h” header file, you can prevent any “pointer from integer” error if you use a function that returns a pointer to “struct _io_file”.
An example of such a function is “fopen()” which allows you to open a file in C programming. Now, the following is a rewrite of the code that causes the pointer error in “struct _io_file”. This time, we use “fopen()” as a pointer to “FILE”.
Now, in the following code, we’ve changed “destination” from an integer to a “character array”. This means “strcpy()” can copy a string into this array without an error.
Meanwhile, the following is another version that turns the “destination” into a “character pointer”. With this, you’ll need to allocate memory using the “malloc()” function from the “stdlib.h” header file.
When you’re assigning pointers, ensure that they’re compatible by changing their data type . The following is the updated code for the “charlie” example , and “qwerty” is now an integer.
This article explained why your code made a pointer without a cast and how you can fix it. The following is a summary of what we talked about:
- An attempt to convert an integer to a pointer will lead to the “makes pointer from integer without a cast wint conversion” error.
- To prevent an “incompatible pointer type” error, don’t set a pointer to a different type.
- If you’re using the “printf()” function, and you want to prevent any “pointer to integer” error, always pass a “format specifier”.
At this stage, you’ll be confident that you can work with integers and pointers in C programming without an error. Save our article, and share it with your developer communities to help them get rid of this error as well.
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Assignment makes pointer from integer without a cast
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Thread: Assignment makes pointer from integer without a cast
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I keep getting this error message: "warning: assignment makes pointer from integer without a cast" for line 17 (which I've made red). I've gotten this message a few times before, but I can't remember how I fixed it, and I can't figure out how to fix this one. Any suggestions? Also, it's a code that will take a password entered by the user and then run several for loops until it matches the password. It prints what it's figured out each time it guesses a new letter. Code: #include <stdio.h> #include <string.h> int main(void) { int i, j; char password[25]; char cracked[25]; char *p; char guess = '!'; printf("Enter a password of 25 characters or less: \n"); scanf("%s", password); printf("Password is being cracked..."); for (i = 0, p = password[i]; i < 25; i++, p++) { for(j = 0; j < 90; j++) { if (*p == guess) { strcpy(p, cracked); printf("\t %s \n"); break; } guess++; } //end <search> for loop } //end original for loop return 0; }
Last edited by deciel; 12-13-2011 at 01:57 AM .
Code: for (i = 0, p = password[i]; i < 25; i++, p++) password[i] is the value at index i of password . You want the address of said value, so you want p = &password[i] (or, equivalently, p = password + i ).
Oh! Thank you, it worked!
Originally Posted by deciel I keep getting this error message: "warning: for (i = 0, p = password[i]; i < 25; i++, p++) [/CODE] Note: password[i] == password[0] == *password since this is the assignment portion of for loop and i is set to zero (0).
If you want to set a pointer to the beginning of an array, just use Code: p = password An array name is essentially a pointer to the start of the array memory. Note, for a null terminated string, you could just test for Code: *p //or more explicitly *p == '\0' Also, a 25-element char array doesn't have room for a 25 character string AND a null terminator. And, ask yourself, what's going on when I enter, say a 10 character password, and i > 10.
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Assignment makes pointer from integer without a cast, warning: assignment makes integer from pointer without a cast, warning: assignment makes integer from pointer without a cast, ' assignment makes pointer from integer without a cast ".
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- warning:assignment makes integer from po
warning:assignment makes integer from pointer without a cast.
C语言assignment makes pointer from integer without a cast
这个警告的意思是将一个int整数值直接赋值给了一个指针变量。( 重点是类型不一致 )
消除警告的方法就是明确类型转换是否是正确的,如果确实要把整数变量赋予指针变量,那么请使用强制类型转换。否则,请用相同的数据类型,这样编译器就不会显示警告。
比如: int *p = 10; //这就会产生这个警告
//因为 p 是指针变量,存放的是地址。而10是一个整数常量
改成: int *p = (int *)10 //强制转换成同一类型就可以消除警告
//强制类型转换,10强制转换成了一个地址
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The routine below draws an image on a tft-screen, and although the whole programm works fine and the image is displayed ok, I'm unable to resolve this 'assignment' warning at pS=pS0+n; .
I've tried several casts, but as yet not the right one (I'm not so well versed in casting).
Can anyone help with the solution?
Definition of the image:
Drawing the image:
The actual routine:
pS0 should have the same type as pS (const char *), and definitely not unsigned int.
TNKernel-PIC32, an open-source real-time kernel for the PIC32
I don't understand why you have ps0 defined as...
If you define it as the correct type, you can avoid all the ugly and unecessary casting:
Edit: andersm beat me to it.
It also seems a lot of your variables are incorrectly sized. rgb8 could easily be unsigned char, for example.
You are trying to convert an pointer to an integer without cast and it is simply telling you that the correct type is what pS type is that is the thing that line is trying to set after all
That all said it would be a hell of a lot easier to make PS0 a const char* and simply tell it
Thank you all for the replies. I think this will help me resolve it. And the the other tips too!
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8 Answers Sorted by: 41 C strings are not anything like Java strings. They're essentially arrays of characters. You are getting the error because strToLower returns a char. A char is a form of integer in C. You are assigning it into a char [] which is a pointer. Hence "converting integer to pointer".
The code is working but it's giving the warning "assignment to 'char' from 'char *' makes integer from pointer without a cast". I need to get rid of this warning. removeCommas (argv [1]); printf ("%s", argv [1]); return 0; for (i=0; i<strlen (num); i++) { c = * (num + i); if (c == 44) { * (num + i) = ""; } } return 0; c Share
Adv Reply March 10th, 2013 #2 r-senior Skinny Extra Sweet Ubuntu Join Date May 2007 Location Leeds, UK Beans 1,675 Distro Ubuntu Re: C - assigment makes integer from pointer without a cast warning path [1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char.
assignment to 'char *' from 'int' makes pointer from integer without a cast. I'm new to C language: include "main.h" int func_d (va_list li) { char *p; int *num; int i, count = 0; num = va_arg (li, int*); p = itoa (num); for (i = 0; p [i] != '\0'; i++) { write (1, &p [i], 1); count++; } return (count); } Then when I compiled this shows up:
We are trying here to assign the return of getinteger function which is a pointer that conatains a memory address to the result variable that is an int type which needs an integer Case 3:...
assignment makes integer from pointer without a castc/c++ warning explained#syntax #c/c++ #compiler #error #warning
Your code made a pointer from an integer without a cast because you assigned an integer to a pointer or you want to convert an integer to a pointer. Other causes include the passing of a variable name to the "printf ()" function and using "struct _io_file in the wrong way. Finally, the following are also possible causes:
The following code in c+ gives me the warning assignment makes integer from pointer without a cast. destination is set as char destination to limit the input string to 10 characters. name is an...
C / C++. 1. warning assignment makes integer from pointer without a cast. by: woods1313drew | last post by: The following code in c+ gives me the warning assignment makes integer from pointer without a cast. destination is set as char destination to limit the input string to 10 characters. name is an...
char *word is a pointer, meaning the value contained by the word parameter is a memory address (in this example, it will be the address of the first character in a string). When you use the notation *word on line 19, you are dereferencing the word pointer, meaning it will return the value of whatever is stored at word's memory address (which is a single character).
1 for (i = 0, p = password [i]; i < 25; i++, p++) password [i] is the value at index i of password. You want the address of said value, so you want p = &password [i] (or, equivalently, p = password + i ). 12-13-2011 #3 deciel Registered User
you can pass search_item as a parameter to the function. That way, value pointed by search_item shall belong to the calling function and will not go out of scope. Also, at least in your present function, search_term does not seem to have any use. You could as well assign the *store pointer, the value of query.
Strings literals are pointers in C. When you a sentence like "Hello world!", that variable is of the type const char*, meaning it is a pointer to a bunch of chars that cannot be modified. When you use double-quotes, you are making a const char*, even if there is only one character. You should use single-quotes if you want to have a char.
1. I am trying to assign a color to the variable choice if it is equal to one of the 3 input numbers. Code: if (pred==1) { choice = "RED"; } else if (pred==2) { choice = "GREEN"; } else if (pred==3) { choice = "BLUE"; } I have already initilized choice as a character variable and pred as an integer variable.
2 Answers Sorted by: 0 In the line arr [MAXLINE * 1000] = strcat (smallVal, largeVal); there are two issues. First, you are not assigning the string to arr, but trying to access the element in position [MAXLINE * 1000], which, in turn is just a plain char. Second, strcat returns an pointer to a char (array), i.e.: char *.
warning: assignment make s integer from pointer without a cas t [enabled by default] C语言 在编译过程中有时候会报警告: warning: assignment make s integer from pointer without a cas t [enabled by default] 这个警告其实不会导致系统运... nota_ assignment 02-25
text should be declared as: char *text = NULL; You need a pointer to char. This pointer can be set to point to any literal string (as you do in your case statements). char text; // this is just a single character (letter), not a string. Objective_Ad_4587 • 3 yr. ago. i got it thank you very much.
You are trying to convert an pointer to an integer without cast and it is simply telling you that the correct type is what pS type is that is the thing that line is trying to set after all. pS = (const char *)(pS0+n) That all said it would be a hell of a lot easier to make PS0 a const char* and simply tell it. pS = (const char *)&pS0[n];
> implicit cast. Explicitly casting these values fixes: > > - warning: assignment to \u2018struct klp_state *\u2019 from \u2018int\u2019 > makes pointer from integer without a cast [-Wint-conversion] > > on lines 38, 55, 68 and 80 of test_klp_state.c I was unable to find where you saw the klp_get_state returning int. I tried
C error - assignment makes integer from pointer without a cast EDIT: SOLVED. Thank you all:) I've tried looking this error up but couldn't understand anything I found, because I am only starting up. I am doing the exercise from C Programming Absolute Beginner's Guide (3rd Edition) and even when printing everything like in the book I get this error.
1 i am new to c, i am trying to make a simple code in which i replace the vowels of an input with the * character... i am having trouble with the array (i am mainly familiar with python only using c now.)
首页 initialization of 'char' from 'char *' makes integer from pointer without a ... \Users\86158\Desktop\c语言\编程练习\第六章02.c [Warning] initialization makes pointer from integer without a cast . ... const reference of type 'int&' from a temporary of type 'int' 这个错误消息表明你在尝试将一个临时 ...