Why does this char array Seg Fault

Can anyone explain to me why when initializing a char array, if the array size is left blank, like this

char str1[] = "Hello";

the program will seg fault, but if it is specified like this

char str1[10] = "Hello";

it works fine.

Here is the full program

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

char concat_string(char str[], char str2[], char destination[], unsigned int bufferSize);

int main(int argc, char *argv[])
{
    unsigned int bufferSize = 64;
    // Both str1 a开发者_运维技巧nd str2 must be defined
    // or else the program will seg fault.
    char str1[] = "Hello ";
    char str2[] = "World";
    char concatenatedString[bufferSize];

    concat_string(str1,str2,concatenatedString,bufferSize);

    printf("The concatenated string is: \n%s\n", concatenatedString);
    return 0;
}


char concat_string(char str[], char str2[], char destination[], unsigned int bufferSize)
{
    char buffer[bufferSize];
    strncat(str, str2, bufferSize);
    strncpy(buffer,str, bufferSize);
    strncpy(destination,buffer,bufferSize);
    return *destination;
}

You have a buffer overflow right here in your concat_string function:

strncat(str, str2, bufferSize);

Your str only has room for seven bytes and it is already full before you try to append str2 to it. You're getting lucky with this:

char str1[10] = "Hello";

as you still don't have enough space allocated to append "World" to it; you're also missing the trailing space on this version of str1 but that's not relevant to your segfault. Your concat_string should be copying str directly to destination and then appending str2 to destination. This would also avoid altering the str and str2 arguments and that would be more polite; you also don't pass the sizes of the str and str1 arrays so there's no way to know if there is room to append anything to them.