Array copying problem

Hi I am trying to copy one struct to another, but I am making some minor mistake that it is not getting copied. Can someone help me fix it?

Code:

---


#include<stdio.h>
#include<stdlib.h>
typedef struct
{
    int inst_addr;
}instruction;

void copy_struct(instruction src,instruction dest)
{
    dest.inst_addr = src.inst_addr;
}

int main()
{
    int addr = 1000;
    instruction check1;
    instruction check2;
    check1.inst_addr = 10;
    check2.inst_addr = 50; 
    copy_struct(check1,check2);
    printf("%d",check2.inst_addr);
}

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You just change the arguments to copy_struct function as follows.

Code:

---

void copy_struct(instruction *src,instruction *dest)
{
    *dest=*src;
}

---

You also do one more change in calling that function as follows.

Code:

---

copy_struct(&check1,&check2);

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Pass the address of the structure.Then inside the function access the member using '->'.Because it is a pointer variable.

Code:

---

#include<stdio.h>
#include<stdlib.h>
typedef struct
{
int inst_addr;
}instruction;

void copy_struct(instruction *src,instruction *dest)
{
    dest->inst_addr = src->inst_addr;
}

int main()
{
    int addr = 1000;
    instruction check1;
    instruction check2;
    check1.inst_addr = 10;
    check2.inst_addr = 50;
    copy_struct(&check1,&check2);
    printf("%d",check2.inst_addr);
}

---

Quote:

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Originally Posted by arun10427

Hi I am trying to copy one struct to another, but I am making some minor mistake that it is not getting copied. Can someone help me fix it?

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As vivekraj suggested, you should change the function to have pointer parameters. However, instead of just assigning pointers, you should assign what the pointers point to, i.e.,

Code:

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*dest = *src;

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EDIT:
That said, this would be rather pointless since you can use it directly. A function like this will come in handy when you need to do a deep copy, e.g., you have a pointer member and you wish to copy what the pointer points to.

Actually we have answered here

Array copying problem