How do I do the following:
int A[2][2];
int B[2];
A[1][0]=2;
B=A[1];
printf("%d",B[0]); //prints out 2
must I use malloc? I know that in higher level languages the above method works, but I'm a little confused about what to do in C
4 Answers
You cannot assign to arrays (though you can initialize them, which is very different).
You can do what you want, by assigning to B element-by-element
B[0] = A[1][0];
B[1] = A[1][1];
or, with a loop
for (k=0; k<2; k++) B[k] = A[1][k];
The problem here is that the variable is declared as int B, when you should declare it as int B[2].
In your code, when you assign A[1] to B, it assigns the value of the first element of A[1] to B
You can declare as int *B, so you will assign the ptr of A[1] to B. As it is, depending on the compiler will not compile. To test, do:
int *B;
B = A[1];
printf("%d, %d", B[0], B[1]);
2 Comments
mugetsu
i made the changes a few moments ago, typo, but the question remains the same as before
Jaison Erick
Above, you will have a reference to A, thus changing B, A is also changed.
You could you memcpy().
#include <stdio.h>
#include <string.h>
int main() {
int a[2][2] = {{1, 2}, {3, 4}};
int b[2];
// Syntax: memcpy(to, from, number_of_bytes_to_copy);
// (to and from should be pointers)
memcpy(b, a[1], sizeof(int) * 2);
printf("%d %d", b[0], b[1]);
}
Output:
3 4
Comments
Arrays are not assignable. You cannot assign the entire array. Whether it's 2D, 1D or 42D makes no difference here at all.
Use memcpy
static_assert(sizeof B == sizeof A[1]);
memcpy(B, A[1], sizeof B);
write your own array-copying function or use a macro
#define ARR_COPY(d, s, n) do{\
size_t i_, n_; for (i_ = 0, n_ = (n); i_ < n_; ++i_) (d)[i_] = (s)[i_];\
}while(0)
ARR_COPY(B, A[1], 2);
answered Oct 3, 2011 at 22:44
AnT stands with Russia
322k44 gold badges548 silver badges793 bronze badges
1 Comment
Mateen Ulhaq
In other words: "You no assign array to array." :)
A[1]is an array with 2 elements; B is an object ot typeint. You cannot assign an array to a single variable: you can assign an array element (of the same type), withB = A[1][0];