string copy function

Code:

---

# include <stdio.h>
# include <String.h>
int main(void){
    int i=0; int j=0;
   
    char *pairA[3];
    char  pairB[10];
   
    pairB[0]='D';
    pairB[1]='o';
    pairB[2]='g';
    pairB[3]='\0'; 

    strcpy(pairA[0],pairB);
}

---

How would I copy the pairB onto pairA ? The program compiles,but does run .

Any other better way to assign the word dog to to pair B ?
would be grateful if some-one can help me out.

Code:

---

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

int main(void){
    char *pairA[3];
    char  pairB[10] = "Dog";

    pairA[0] = malloc(strlen(pairB)+1);
    strcpy(pairA[0], pairB);
    free(pairA[0]);
}

---

[edit]
Dynamic memory might not be something a beginner should start out with, an easier way would be to make pairA a 2D array instead of a an array of char pointers.

Code:

---

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

int main(void){
    char pairA[3][10];
    char  pairB[10] = "Dog";

    strcpy(pairA[0], pairB);
}

---

Quote:

---

Originally Posted by rahulsk1947

[
Any other better way to assign the word dog to to pair B ?

---

Well, you could simply make pairA a pointer to your string.

Code:

---

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

int main() {
    char  *pairA;
    char  pairB[10];

    pairB[0]='D';
    pairB[1]='o';
    pairB[2]='g';
    pairB[3]='\0'; 

    pairA = pairB;
   
    printf("%s", pairA);
}

---

Ofcourse if pairB changes... so does pairA.

Technically, 'pairA' doesn't change. The value it contains is still the address of 'pairB'. It appears to change if you dereference it, because the value of what it points to has changed. ;)


Quzah.

Quote:

---

Originally Posted by rahulsk1947

How would I copy the pairB onto pairA ? The program compiles,but does run .

Any other better way to assign the word dog to to pair B ?
would be grateful if some-one can help me out.

---

I'd go with something like this.

Code:

---

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

int main(void)
{
  char pairB[] = "Dog", pairA[sizeof pairB];
  strcpy(pairA, pairB);
  /* do stuff */
  return 0;
}

---

Hmm,so

Code:

---

char pairB[10]="dog";

---

is the way to assign it.

infact I been trying to do

Code:

---

char pairB[10];  pairB="dog";

---

but i get compilation errors.

actually i am trying to do

Code:

---

#include <stdio.h>

struct student{
    char name[25];
    int age;
    int idNum;
};

int main(void){
    struct student a;
   
    a.age=21;
    a.idNum=77273869;
    // a.name="rahul"; -> gives an error
   
}

---

how to assign my name to a.name ??

Code:

---

memcpy(a.name,"rahul",strlen(a.name)+1);  // include string.h>

or

strcpy(a.name,"rahul");  // < string.h >

---

ssharish2005

You can't make assignments to entire arrays except at declaration time. You must do a cell at a time. You can use the standard library function strcpy to do this for you. All it does is loop through one string, copying it to another. Dave has already showed you how to use it up above.


Quzah.