Converting from integer to string in C

Question

I want to convert an integer to a string in C. I have tried the following code but the program is constantly outputting a 9-digit number. Does someone knows what is the error please and how can I fix it?

int num = 158;
char str[5];
sprintf(str, "%d" ,num);
printf("The result is: %d" , sprintf);

Thanks!

Well, you want to print `str` (not `sprintf`) with the `%s` (not `%d`) format. Doesn't your compiler warn you about the `sprintf`? — M Oehm, Nov 13 '14 at 07:48
Also, 4 chars (plus null terminator) may be enough in your case, but it isn't for arbitrary `int`s. Might want to adjust the buffer size to 12 (to cater for -2147483648) and also use `snprintf` to avoid buffer overflows. — M Oehm, Nov 13 '14 at 07:50
you're printing the address of sprintf, typically a very large number, and you're also printing using the wrong format which results in undefined behavior — phuclv, Nov 13 '14 at 07:52

score 6 · Answer 1 · answered Nov 13 '14 at 07:56

Either you print the integer with %d

printf("The result is: %d\n", num);

or the string representation with %s

printf("The result is: %s\n" , str);

By doing

printf("The result is: %d" , sprintf);

You are printing the decimal representation of the address of the function sprintf. Example:

#include <stdio.h>
int main() {
int num = 158;
char str[5];
sprintf(str, "%d" ,num);
printf("The result is: %d\n", sprintf);
printf("The result is: %8x\n", sprintf);

}

Compile statically in order to make it easier to locate the address of sprintf.

➜  ~ [4] [Thu 13] $ gcc file.c -o bin -static

In the code, I also print the hexadecimal representation, which is easier to locate in the binary file. Output:

The result is: 4200768
The result is:   401940

You can actually check the linear address of sprintf in the ELF executable:

➜  ~ [4] [Thu 13] $ nm bin | grep sprintf
0000000000480830 W asprintf
0000000000480830 T __asprintf
0000000000480830 T ___asprintf
0000000000401940 T _IO_sprintf
0000000000480a40 T _IO_vasprintf
00000000004019d0 T __IO_vsprintf
00000000004019d0 T _IO_vsprintf
0000000000401940 T sprintf
0000000000401940 T __sprintf
0000000000480a40 W vasprintf
00000000004019d0 W vsprintf

As expected, 0x0000000000401940.

`As expected...` No it is UB, I did not expect anything. There should be no complaining even if the [computer blows up](http://stackoverflow.com/questions/26882023/is-it-good-idea-to-pass-uninitialized-variable-to-srand/26882097#26882097). — Mohit Jain, Nov 13 '14 at 09:38

score 3 · Answer 2 · answered Nov 13 '14 at 07:47

3

Change your printf to printf("The result is: %s" , str);.

%s is the specifier for strings and your string name is str. Printing with incorrect % specifiers invoke undefined behavior.

answered Nov 13 '14 at 07:47

Mohit Jain

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David Heffernan · Answer 3 · 2014-11-13T08:07:21.743

printf("The result is: %d" , sprintf);

This code attempts to print sprintf which is a function. So its address is passed to printf. That's just not what you intended to do. Not to mention the fact that %d with an address leads to undefined behavior.

To print the string you made, you do this:

printf("The result is: %s", str);

Note that you must use the %s format string because the argument you supply is a string.

If all you want to do is to print the value, then you can remove str, remove the call to sprintf, and get printf to perform the formatting:

printf("The result is: %d", num);

One advantage of this is that it avoids you having to decide how large a buffer to allocate. You allocated a buffer with length 5 which can accept numbers up to 4 digits, or 3 digits if negative. For values with more digits, then your code will overrun that buffer.

Converting from integer to string in C

3 Answers3