integer to string method [duplicate]

As it has already been mentioned in comments and other answers, returning a local variable (aka buffer) is something that you shall never do as local variables are destroyed once the function return.

Further you buffer is too small to hold the 4 characters as a string in C requires an extra character to zero terminate the string. So to hold 4 character you'll (at least) need buffer[5]. However, notice that %04d does not ensure that there will be printed exactly 4 characters. A high int value will produce more characters and result in (more) buffer overflow. So you'll need a buffer that can hold a print of the largest (perhaps negative) integer.

So what can you do instead?

You have two options. 1) Use dynamic memory allocation inside the function or 2) let the caller supply the destination buffer to the function.

It may look something like:

#include <stdio.h> #include <stdlib.h> // The maximum number of chars required depends on your system - see limits.h // Here we just use 64 which should be sufficient on all systems #define MAX_CHARS_IN_INT 64 char* intToMallocedString(int Value) { char* buffer = malloc(MAX_CHARS_IN_INT); // dynamic memory allocation sprintf(buffer, "%04d", Value); return buffer; } // This could also be a void function but returning the buffer // is often nice so it can be used directly in e.g. printf char* intToProvidedString(char* buffer, int Value) { sprintf(buffer, "%04d", Value); return buffer; } int main(void) { int x = 12345678; char str[MAX_CHARS_IN_INT]; // memory for caller provided buffer char* pStr = intToMallocedString(x); intToProvidedString(str, x); printf("%s - %s\n", str, pStr); free(pStr); // dynamic memory must be free'd when done return 0; } 

Output:

12345678 - 12345678