Why do i get this stack smashing error in my program? - c

Question

I am practicing my use of strcpy() in C and wrote this program.
#include <stdio.h>
#include <string.h>
int main(){
int array[4] = {3,7,1,5};
char str[4], temp[5];
for (int i = 0; i < 4; i++){
sprintf(&str[i],"%d", array[i]);
}
for (int i = 0; i < 4; i++){
strcpy(&temp[i],&str[i]);
printf("%c", temp[i]);
}
return 0;
}
~
Unfortunately i get a 'stack smashing' error why is this ?

Answer 1

Strings in C must be terminated by a NUL byte ('\0'), and you must account for that when you size your buffers.
Given the contents of array, the following code
for (int i = 0; i < 4; i++) {
sprintf(&str[i], "%d", array[i]);
}
will write bytes as follows, remembering that arrays are indexed from 0:
'3' to the 0th index, '\0' to the 1st index
'7' to the 1st index, '\0' to the 2nd index
'1' to the 2nd index, '\0' to the 3rd index
'5' to the 3rd index, '\0' to the 4th index
You write over the previously placed NUL with the current integer. Do you see the problem, though? There is not a 4th index in str[4]. You're going to write a byte to memory that you should not be accessing.
Even with str[5], so that there is room for the final NUL, the second part of your code would work in a very counter productive way.
The results of the first loop would have the contents of str be equal to "3714", such that
strcpy(&temp[i], &str[i]);
would then copy the strings as such:
"3714" to temp + 0
"714" to temp + 1
"14" to temp + 2
"4" to temp + 3
See how you're just walking along the same string, recopying the same tails to the same locations? printf("%c", temp[i]); is just going to print the first character of whichever tail you're on.
With enough room in str, this program is essentially a very round about way of writing:
int array[] = { 3, 7, 1, 5 };
for (int i = 0; i < 4; i++)
printf("%d", array[i]);
Additionally, if array were to contain any integers whose representation as a string required more than a single byte (e.g., -5, 32), this code would not scale at all.
You'll want to change str to be a 2d array, containing space for all the strings you will create from the integers in array
#include <stdio.h>
#include <string.h>
int main(void) {
int array[4] = {3, 7, 1, 5};
char str[4][12], temp[12];
for (int i = 0; i < 4; i++)
sprintf(str[i], "%d", array[i]);
for (int i = 0; i < 4; i++){
strcpy(temp, str[i]);
printf("[%d] %s\n", i, temp);
}
}
though this is still not a very interesting exercise for the use of strcpy, seeing as we could just print str[i] directly.
Here's an example to play around with of using strcpy when searching for the last integer in an array whose string representation is longer than than two characters.
#include <stdio.h>
#include <string.h>
int main(void) {
int array[] = { 3, 7, 123, 4, 13, -55, 29, 55 };
char string[12] = "0", temp[12];
for (int i = 0; i < (sizeof array / sizeof *array); i++)
if (sprintf(temp, "%d", array[i]) > 2)
strcpy(string, temp);
printf("Final value: %s\n", string);
}

Answer 2

Each str[i] can only hold a single character, not a string - when you write
sprintf( &str[i], "%d", array[i] );
you are attempting to write a 2-character string1 to a 1-character buffer; the string terminator is "spilling" into the array element str[i+1] until you get to str[3], in which case the terminator spills over the end of the array.
To store an array of strings, you need to set aside a 2D array of char:
char str[4][N+1]; // can hold 4 strings of length N, +1 for the string terminator
Then in your sprintf statement, you'd write
sprintf( str[i], "%d", array[i] ); // no & operator on str[i]
All the values in array are only one digit - for larger values, you would need more characters in your array. A string representation of a 32-bit int can have up to 10 digit characters, plus a sign character, plus the string terminator, so you need to set aside an array that can hold 12 elements, so ideally str should be declared as char str[4][12];