CODEWITHSUNDEEP
carraysassign
Morfeo Rejocktana
Morfeo Rejocktana

Reputation: 9

how to assign char array to another char array with another size in C

I want to assign an array to another array. Both arrays do not have the same size:

char array1[7] = "abcdefg"; 
char array2[3];

How can i assign three(3) values from array1 into array 2? I tried so:

array2[3] = {array1[2], array1[3], array1[4]};

and i´d become the following error message:

expected expression before ´{´token

I know that i can use (memcpy()) when the arrays have the same size.

I have more experience with VHDL and verilog. In VHDL it could look like this:

array2 := arry1(2 upto 4);

But i do not have much experience with C programing.

Thank you

Upvotes: 0

Views: 7370

Answers (4)

Ilmari Karonen
Ilmari Karonen

Reputation: 50328

If you wish to do the copy at run-time, you can either use memcpy():

memcpy(array2, array1 + 2, 3);

or just write your own copy loop:

for (int i = 0; i < 3; i++) array2[i] = array1[i + 2];

Note that, depending on the surrounding code and your C compiler and optimization settings, these two methods may or may not end up compiling into the same binary. Specifically, it's quite possible for an optimizing compiler to both inline memcpy() and to transform the explicit copy loop into a call to memcpy(), depending on what the compiler thinks would be most efficient.

Also note that, if the arrays were of some type other than char, the length argument to memcpy() would need to be multiplied by the size of the element type in bytes. However, the C standard guarantees that sizeof(char) == 1.

Upvotes: 2

Vlad from Moscow
Vlad from Moscow

Reputation: 310930

If you indeed mean the assignment then you can use еру standard function strncpy declared in header <string.h>.

Here is a demonstrative program

#include <stdio.h>
#include <string.h>

int main(void) 
{
    char array1[7] = "abcdefg"; 
    char array2[4];
    size_t n = 3;

    strncpy( array2, array1 + 2, n );
    array2[n] = '\0';

    puts( array2 );

    return 0;
}

Its output is 

cde

Take into account that if you want that the array array2 would contain a string then its size shall be equal to 4 to store the three copied characters and the terminating zero.

Upvotes: 1

Marichyasana
Marichyasana

Reputation: 3154

This works on Windows 7 

#include <stdio.h>
#include <string.h>

int main() {
    char array1[7] = "abcdefg"; 
    char array2[3];
    strncpy(&array2[0],&array1[2],3);
    printf("%s\n",array2);
}

Output:

cde

Upvotes: 0

Berkay92
Berkay92

Reputation: 588

When you initialize your array2[3], you can copy the values like you do

Just do this,

char array1[7] = "abcdefg"; 
char array2[3] =  {array1[2], array1[3], array1[4]};

Upvotes: 2

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