How to reassign a array in c? - c

Question

#include <stdio.h>
#include <stdlib.h>
void reasi(char** a){
char* x[] = {"1","22","333"};
a = x;
}
int main(){
char* a[] = {"bob","alice","tom"};
reasi(a);
for(int i=0; i< 3; i++){
printf("%s\n",a[i]);
}
}
The desired output should be {"1","22","333"}, but it won't work if I assign the value like that. I do know how to change the value of an int or char but don't know how to reassign the value to an array (without dynamically allocating memory). I tried to update each element inside "a" and it works. Thank you.

Answer 1

What you're doing won't work. You're simply creating a local array and then assigning your local parameter a to the beginning of this array (which changes nothing about the a in main). So the real thing isn't modified.
To actually modify this, you can either do a plain for loop:
// NOTE: this assumes array has the same number of elements as x
void reasi(char** a)
{
char* x[] = {"1","22","333"};
for (unsigned i = 0; i < sizeof x / sizeof *x; ++i)
a[i] = x[i];
}
Or use memcpy:
#include <string.h>
// NOTE: this assumes array has the same number of elements as x
void reasi(char** a)
{
char* x[] = {"1","22","333"};
memcpy(a, x, sizeof x);
}

Answer 2

Your array is not a 2D array. By seeing your code I assume you mistaken 1D array for 2D array, hence I will answer according to it
#include <stdio.h>
#include <stdlib.h>
void reasi(char** a){
a[0] = "1";
a[1] = "22";
a[2] = "333";
}
int main()
{
char* a[] = {"bob","alice","tom"};
reasi(a);
for(int i=0; i< 3; i++)
{
printf("%s\n",a[i]);
}
}
This will give you your desired output.

Answer 3

In C an array has a fixed size. You cannot resize it after the fact. If the size always stays the same, you can copy the new array:
#include <stdio.h>
#include <string.h>
void reasi(char const **a) {
char const *x[] = {"1", "22", "333"};
memcpy(a, x, sizeof x);
}
int main() {
char const *a[] = {"bob", "alice", "tom"};
reasi(a);
for (int i = 0; i < 3; i++)
puts(a[i]);
}
If you do want to resize the array, you are going to have to allocate it dynamically with malloc.

Answer 4

Both a and x are each an array of pointers to char. In C, you cannot assign the contents of an array C11 Standard - 6.3.2.1 Other Operands - Lvalues, arrays, and function designators(p3) "the array object ... is not an lvalue."
Instead, you must assign each element (pointer) from x to a or use memcpy() to accomplish the same.
Further, hardcoding the contents of x in your function makes little sense. Why? You have just written a function that does nothing but assign the pointers (to String-Literals "1", "22", "333") and is incapable of doing anything else -- useful.
Why not declare x in main() and pass it as a parameter along with a and the number of elements? That way, you can pass any array of pointers to char as x (with at least 3 elements) and reassign the elements to a).
For example:
#include <stdio.h>
void reasi (char **a, char **x, size_t nelem)
{
for (size_t i = 0; i < nelem; i++) {
a[i] = x[i];
}
}
int main() {
char *a[] = {"bob","alice","tom"},
*x[] = {"1","22","333"},
*y[] = {"4","55","666","7777","8888"};
size_t n = sizeof a / sizeof *a;
reasi (a, x, n);
puts ("x->a");
for (size_t i = 0; i < n; i++) {
printf ("%s\n", a[i]);
}
reasi (a, y, n);
puts ("\ny->a");
for (size_t i = 0; i < n; i++) {
printf ("%s\n", a[i]);
}
reasi (a, y + 2, n);
puts ("\ny+2->a");
for (size_t i = 0; i < n; i++) {
printf ("%s\n", a[i]);
}
}
The refactoring above generalizes your reasi() function, making it reusable and a bit more useful than a single use case of "1", "22", "333".
Example Use/Output
Running you get the expected:
$ ./bin/reasi
x->a
1
22
333
y->a
4
55
666
y+2->a
666
7777
8888
Wrapping memcpy() in a function in that case wouldn't buy you any benefit, you could simply call memcpy (a, x, n * sizeof *a); from main() and avoid the function call overhead (which a decent compiler would likely optimize out anyway).
Look things over and let me know if you have further questions.