C Dynamic char array - c

Question

Expanding array dynamicly after user enter string in function dynamic_array My Problem seems to be when i try to use the extended array agian in main after i dynamic_array returns true.
After function call i try to print input with printf("main string: %s\n", input) the program will crash. It seems like the *input in main never gets extended.
int dynamic_array(char *input, int *string_current_len){
int string_len = 0;
char temp_c;
input = (char *) malloc(sizeof(char));
if(input == NULL) {
printf("Could not allocate memory!");
exit(1);
}
printf("File to search in: ");
while((temp_c = getchar()) != '\n') {
realloc(input, (sizeof(char)));
input[string_len++] = temp_c;
}
input[string_len] = '\0';
printf("\nYou entered the string: %s\n", input);
printf("length of string is %d.\n", string_len);
*string_current_len = string_len;
return 1;
}
int main(void) {
int string_len = 0;
char *input;
printf("enter #q as filename or word to quit.\n");
if(!dynamic_array(input, &string_len)){
return 0;
}
printf("main string: %s\n", input);
return 0;
}

Answer 1

This:
realloc(input, (sizeof(char)));
is wrong. The realloc() function doesn't modify the given pointer (it can't!), it returns the new pointer. It can also fail, and return NULL.
Also, the second argument doesn't make any sense at all, it should be the new desired total size of the previously allocated buffer, but you're always passing (a very obscure) 1. It's not "grow this by this amount", it's the rather more low-level "attempt to grow this to this new size, and return the new location of the grown buffer".
Please read the documentation very carefully.

Answer 2

realloc(input, (sizeof(char)));
You are reallocating with same size (i.e 1 byte). It shoud be:
while((temp_c = getchar()) != '\n') {
realloc(input, (string_len + 1) * (sizeof(char)));
input[string_len++] = temp_c;

Answer 3

if(!dynamic_array(input, &string_len)){
return 0;
}
"input" variable is used without initialization.
realloc(input, (sizeof(char)));
Above "realloc" is returning bad pointer. It may be totally bogus, or it may have been allocated from another heap. The pointer MUST come from the 'local' heap.

Answer 4

C has a call-by-value semantics. So any changes to formal input inside dynamic_array is not propagated to the caller (e.g. your main).
Your main does not initialize input. If you compiled with all warnings and debug info (as you should), e.g. with gcc -Wall -g, you'll get a warning about that.
I actually recommend to initialize every local variable. This makes the debugging easier (since runs are more reproductible). Useless initializations will be removed by the optimizer.
You could initialize input inside your main with
char* input = NULL;
and you should redesign your program, perhaps by having a grow_array function (instead of your dynamic_array) which you would call in your main like
grow_array (&input, &string_len);
I leave up to you to declare and implement grow_array correctly. I'm too lame to do your homework.
Inside your grow_array you want to call malloc and test it:
*pptr = malloc(newsize);
if (!*pptr) { perror ("malloc"); exit (EXIT_FAILURE); };
Don't forget to use the debugger (gdb) to run your program step by step.
I don't recommend using realloc because error handling could be tricky. I would suggest using malloc and free and cleverly copying the data using mempcy.
Read the documentation of every function that you are using, e.g. malloc(3), printf(3)

Answer 5

couple of things that I noticed.
int dynamic_array(char *input, int *string_current_len) should change to
int dynamic_array(char **input, int *string_current_len)
since this function is trying to modify a pointer.
also the call to the function here
if(!dynamic_array(input, &string_len)){
return 0;
}
needs to be:
if(!dynamic_array(&input, &string_len)){
return 0;
}