Question

I have never really done much C but am starting to play around with it. I am writing little snippets like the one below to try to understand the usage and behaviour of key constructs/functions in C. The one below I wrote trying to understand the difference between char* string and char string[] and how then lengths of strings work. Furthermore I wanted to see if sprintf could be used to concatenate two strings and set it into a third string.

What I discovered was that the third string I was using to store the concatenation of the other two had to be set with char string[] syntax or the binary would die with SIGSEGV (Address boundary error). Setting it using the array syntax required a size so I initially started by setting it to the combined size of the other two strings. This seemed to let me perform the concatenation well enough.

Out of curiosity, though, I tried expanding the “concatenated” string to be longer than the size I had allocated. Much to my surprise, it still worked and the string size increased and could be printf‘d fine.

My question is: Why does this happen, is it invalid or have risks/drawbacks? Furthermore, why is char str3[length3] valid but char str3[7] causes “SIGABRT (Abort)” when sprintf line tries to execute?

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void main() {
    char* str1 = "Sup";
    char* str2 = "Dood";

    int length1 = strlen(str1);
    int length2 = strlen(str2);
    int length3 = length1 + length2;

    char str3[length3];
    //char str3[7];

    printf("%s (length %d)\n", str1, length1);           // Sup (length 3)
    printf("%s (length %d)\n", str2, length2);           // Dood (length 4)
    printf("total length: %d\n", length3);               // total length: 7
    printf("str3 length: %d\n", (int)strlen(str3));      // str3 length: 6
    sprintf(str3, "%s<-------------------->%s", str1, str2); 
    printf("%s\n", str3);                                // Sup<-------------------->Dood

    printf("str3 length after sprintf: %d\n",            // str3 length after sprintf: 29
            (int)strlen(str3));
}

Answer 1

void main() {
    char* str1 = "Sup"; // a pointer to the statically allocated sequence of characters {'S', 'u', 'p', '\0' }
    char* str2 = "Dood"; // a pointer to the statically allocated sequence of characters {'D', 'o', 'o', 'd', '\0' }

    int length1 = strlen(str1); // the length of str1 without the terminating \0 == 3
    int length2 = strlen(str2); // the length of str2 without the terminating \0 == 4
    int length3 = length1 + length2;

    char str3[length3]; // declare an array of7 characters, uninitialized

So far so good. Now:

printf("str3 length: %d\n", (int)strlen(str3));      // What is the length of str3? str3 is uninitialized!

C is a primitive language. It doesn’t have strings. What it does have is arrays and pointers. A string is a convention, not a datatype. By convention, people agree that “an array of chars is a string, and the string ends at the first null character”. All the C string functions follow this convention, but it is a convention. It is simply assumed that you follow it, or the string functions will break.

So str3 is not a 7-character string. It is an array of 7 characters. If you pass it to a function which expects a string, then that function will look for a '\0' to find the end of the string. str3 was never initialized, so it contains random garbage. In your case, apparently, there was a '\0' after the 6th character so strlen returns 6, but that’s not guaranteed. If it hadn’t been there, then it would have read past the end of the array.

sprintf(str3, "%s<-------------------->%s", str1, str2); 

And here it goes wrong again. You are trying to copy the string "Sup<-------------------->Dood\0" into an array of 7 characters. That won’t fit. Of course the C function doesn’t know this, it just copies past the end of the array. Undefined behavior, and will probably crash.

printf("%s\n", str3);  // Sup<-------------------->Dood

And here you try to print the string stored at str3. printf is a string function. It doesn’t care (or know) about the size of your array. It is given a string, and, like all other string functions, determines the length of the string by looking for a '\0'.