C Array to Pointers

Question

Please explain why the below code fails at line 10. How can I print values of p, i.e. Hello World.

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

int main(int argc, char **argv) {
    char ar[200] = "Hello World";
    strcat(ar, " !");
    printf("%s\n", ar);

    char **p = &ar;
    printf("%s\n", *p[0]);

    return 0;
}

EDIT:

Just to clarify more on what I want to achieve. I have a function that accepts char** as argument, hence why I want to convert char ar[200] into char**, but the application freezes.

It's either printf("%s\n", p[0]); or printf("%s\n", *p);. By doing printf("%s\n", *p[0]);, the pointer to string p is dereferenced twice. Hence, *p[0] is a char, not a string. — francis
– francis, Commented May 20, 2016 at 11:40
@RoshDonniet, I have tried that but the program crashes at the end — manaila
– manaila, Commented May 20, 2016 at 11:42
The type of &ar is char (*)[200] (pointer to 200-element array of char), not char **. — John Bode
– John Bode, Commented May 20, 2016 at 11:42

Anish Sharma · Accepted Answer · 2016-05-20 12:00:46Z

5

You have declared a character array and assigned value to it. If you just mention the name of the array, you are actually mentioning the base address of this array. Pointers can store address, you can define a char pointer and assign the base address of your character array to it.

for example

char ar[200]="Hello World";
char *p=a; //assign the base address of a to p;

You can then print the string using the %s format specifier with the base address of your character array(string).

printf("%s",a);// print the string "Hello World". here a is the base address
printf("%s",p);//print the string "Hello World". here p is the base address

This will work

#include <stdio.h>
#include <string.h>
int main(int argc, char **argv) {
    char ar[200] = "Hello World";
    strcat(ar, " !");
    printf("%s\n", ar);
    char *p = ar;
    printf("%s\n", p);
    return 0;
}

Edited for your convenience

char *q=a;
char **P=&q;
printf("%s",*p);

if this is what you want

edited May 20, 2016 at 12:00

answered May 20, 2016 at 11:38

Anish Sharma

5253 silver badges17 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Rosh Donniet Over a year ago

You should explain why it works and his code doesn't

Anish Sharma Over a year ago

@JohnColeman I appreciate your world and would really like to thank him

Peter - Reinstate Monica · Accepted Answer · 2016-05-20 15:45:03Z

Arrays already "decay" to the address of their first element in most contexts. This behavior differs from other variables, even from the other aggregates, structs. The reasons are rooted deep in the history of C (namely, no joke, in B¹).

One of the contexts though where an array does not decay is when its address is taken: the address operator yields a pointer to array, not a pointer to pointer to char:

char arr[6] = "abcde";  // 5 letters plus '\0'
char (*parr)[6] = &arr; // pointer to array of 6 elements

Curiously enough, dereferencing this pointer twice yields the first char in the array, as if parr were a pointer to pointer to char. The reason is that *parr simply is the original array, and dereferencing that is quite normally its first element. This is likely the source of much confusion:

printf("%c", **parr);

Traditionally (i.e. in the 1970s) pointer was pointer, and their types didn't matter much. (There is a big difference though: What value does parr+1 have?) But it's nice to compile without warnings and to understand what one is doing. So how do we get your requirement, a pointer to a pointer to a char, for printing arr? The answer is, just use a pointer to a pointer!

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

int main(int argc, char **argv) {
    char ar[200] = "Hello World";
    strcat(ar, " !");
    printf("%s\n", ar);

    char *p = ar;  // pointer to char: 
                   // array "decays" to address of first char

    char **pp = &p;  // pp is a pointer to a pointer to char.
                     // The point is, it points to the "pointer to char" p.

    printf("%s\n", *pp); // *pp is what pp points to: 
                         // p, a simple pointer to char. That's what "%s"
                         // expects. p is initialized and points to 
                         // the first char in ar.

    return 0;
}

¹Cf. the interesting paper by Dennis Ritchie, "The Development of the C Language", p.7

yayg · Accepted Answer · 2016-05-20 11:49:53Z

1

gcc     main.c   -o main
main.c: In function ‘main’:
main.c:9:16: warning: initialization from incompatible pointer type [enabled by default]
     char **p = &ar;
                ^
main.c:10:5: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=]
     printf("%s\n", *p[0]);
     ^

expects argument of type ‘char *’, but argument 2 has type ‘int’ seems very clear :P

You can print your string by changing you program by

int main(int argc, char **argv) {
    char ar[200] = "Hello World";
    strcat(ar, " !");

    char *p = ar;
    printf("%s\n", p);

    return 0;
}

edited May 20, 2016 at 11:49

answered May 20, 2016 at 11:44

yayg

3192 silver badges12 bronze badges

Comments

Viktor Simkó · Accepted Answer · 2016-05-20 12:10:00Z

1

In c ar is equal to &ar, because arrays are basically memory addresses to their first elements.
So you only need a pointer to a char not a pointer to a pointer to a char.

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

int main(int argc, char **argv) {
    char ar[200] = "Hello World";
    strcat(ar, " !");
    printf("%s\n", ar);

    char *p = &ar;
    printf("%s\n", p);

    return 0;
}

edited May 20, 2016 at 12:10

answered May 20, 2016 at 11:43

Viktor Simkó

2,63718 silver badges22 bronze badges

Comments

John Coleman · Accepted Answer · 2016-05-20 12:36:39Z

1

This question is an interesting example of the difference between arrays and pointers. It is common to think of arrays as pointers, but that isn't quite true. The fact that char ar[200] doesn't make ar into a char pointer is why you can't use a char** to point to ar. On the other hand, arrays decay to pointers in function calls, so the following code does work (though it is terribly unmotivated):

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

void printString(char ar[]){
    char **p = &ar; // as in original code
    printf("%s\n",*p);
}

int main(int argc, char **argv) {
    char ar[200] = "Hello World";
    strcat(ar, " !");
    printf("%s\n", ar);

    printString(ar); 

    return 0;
}

edited May 20, 2016 at 12:36

answered May 20, 2016 at 12:10

John Coleman

52.1k7 gold badges59 silver badges127 bronze badges

Comments

Rishikesh Raje · Accepted Answer · 2016-05-20 11:49:34Z

0

How can I print values of p, i.e. Hello World.

You do not need a pointer to pointer for this purpose. A simple pointer will do.

char *p;
p = ar;
printf ("%s\n",p);

answered May 20, 2016 at 11:49

Rishikesh Raje

8,6122 gold badges18 silver badges33 bronze badges

Comments

Stian Skjelstad · Accepted Answer · 2016-05-20 12:32:41Z

-1

Example of when to use **pointers

#include <stdio.h>

void foo(char **bar)
{
  *bar = "some text";
}

int main(int argc, char *argv[])
{
  char *temp;

  foo (&temp);

  printf ("temp=%s\n", temp);

  return 0;
}

answered May 20, 2016 at 12:32

Stian Skjelstad

2,3451 gold badge12 silver badges20 bronze badges

Comments

Ronaldo Felipe · Accepted Answer · 2016-05-20 17:12:30Z

-2

I explain in the comments of your code.

EDIT: Do not copy following code

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

int main(int argc, char **argv) {
    char ar[200] = "Hello World";// technically an array is a pointer, so char *ar and char[] are the same
    strcat(ar, " !");// pass the pointer to treat strings
    printf("%s\n", ar);// pass the pointer to treat strings

    char **p;// here is the problem you were giving the address of your pointer to a pointer of pointer in inline wich makes more confusing
    (*p) = &ar; // here is the correct assingment
    printf("%s\n", *p);// so here you should pass  *p
    return 0;
}

EDIT: This solution is far safer in real life situations

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

int main(int argc, char **argv) {
    char ar[200] = "Hello World";// technically an array is a pointer, so char *ar and char[] are the same
    strcat(ar, " !");// pass the pointer to treat strings
    printf("%s\n", ar);// pass the pointer to treat strings

    // This solution initializing is better because it avoids editing other variables
    char *p = ar;
    char **pp = &p;

    printf("%s\n", *pp);// so here you should pass  *pp
    return 0;
}

edited May 20, 2016 at 17:12

answered May 20, 2016 at 11:45

Ronaldo Felipe

4022 silver badges13 bronze badges

11 Comments

Stian Skjelstad Over a year ago

(*p) = &ar; this will dereference p, which is not initialized

Ronaldo Felipe Over a year ago

A pointer doesn't need to be initialized because it points to an address in memory. Here is the test working properly: (ideone.com/cnM9Jn#stdin)

Stian Skjelstad Over a year ago

gcc complains: test95.c:10:10: warning: ‘p’ is used uninitialized in this function [-Wuninitialized]

Anish Sharma Over a year ago

can you cite the difference between ar and &ar??

Peter - Reinstate Monica Over a year ago

I have programmed a proof of concept how your code overwrites another variable in the program. Cf. coliru.stacked-crooked.com/a/2a9c14e9bae022ac. This kind of errors is hard to find.

|

Collectives™ on Stack Overflow

C Array to Pointers

8 Answers 8

Edited for your convenience

2 Comments

Comments

Comments

Comments

Comments

Comments

Comments

11 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

8 Answers 8

Edited for your convenience

2 Comments

Comments

Comments

Comments

Comments

Comments

Comments

11 Comments

Your Answer

Sign up or log in

Post as a guest

Related