Question

Here is my code:

    printf("%s\n", "test1");
    char c = '2';
    char * lines[2];
    char * tmp1 = lines[0];

    *tmp1 = c;
    printf("%s\n", "test2");

I don’t see the second printf in console.

Question: Can anybody explain me what’s wrong with my code?

NOTE: I’m trying to learn C 🙂

Answer 1

This line:

char * lines[2];

declares an array of two char pointers. However, you don’t actually initialize the pointers to anything. So later when you do *tmp1 = (char)c; then you assign the character c to somewhere in memory, possibly even address zero (i.e. NULL) which is a bad thing.

The solution is to either create the array as an array of arrays, like

char lines[2][30];

This declares lines to have two arrays of 30 characters each, and since strings needs a special terminator character you can have string of up to 29 characters in them.

The second solution is to dynamically allocate memory for the strings:

char *lines[2];
lines[0] = malloc(30);
lines[1] = malloc(30);

Essentially this does the same as the above array-of-arrays declaration, but allocates the memory on the heap.

Of course, maybe you just wanted a single string of a single character (plus the terminator), then you were almost right, just remove the asterisk:

char line[2];  /* Array of two characters, or a string of length one */