无法将char分配给char指针

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 (ie 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 */

Answer 2

lines is uninitialized, and tmp1 initialization is wrong.

It should be:

char lines[2];
char * tmp1 = lines;

Alternatively, you can say:

char * tmp1 = &lines[0];

Or else for an array of strings:

char lines[2][30];
char * tmp1 = lines[0];

Answer 3

The array lines in uninitialized. Thus lines[0] is an uninitalized pointer. Dereferencing it in *tmp1 is sheer undefined behaviour .

Here's an alternative, that may or may not correspond to what you want:

char lines[2];
char * tmp1 = lines;   // or "&lines[0]"

*tmp = c;

Or, more easily:

char lines[2] = { c, 0 };

Answer 4

The line

char * lines[2];

creates an array of two char pointers. But that doesn't allocate memory, it's just a "handle" or "name" for the pointer in memory. The pointer doesn't point to something useful.

You will either have to allocate memory using malloc() :

char * lines = malloc(2);

or tell the compiler to allocate memory for you:

char lines[2];

Note: Don't forget to terminate the string with a 0 byte before you use it.

Answer 5

char *lines[2]; : A two element array of char pointers.

char *tmp; : A pointer to a character.

char *tmp = lines[0] : The value inside the array element 0 of the lines array is transferred into tmp . Because it is an automatic array, therefore it will have garbage as the value for lines[0] .

*temp : Dereference the garbage value. Undefined Behaviour.

Answer 6

char * tmp1 = lines[0];

here you declare a char pointer and initialize its pointer value to line[0],the fist element stored in line array which is also uninitialized.

now the tmp is pointing to somewhere unknown and not actually accessible. When you

*tmp1 = (char)c;

you are operating on a invalid memory address which causes a Segmentation fault.