CODEWITHSUNDEEP
cpointers
mko
mko

Reputation: 22064

Why the type doesn't match in this value swap code in C?

void swap(char *a,char *b){
  char t;
  t = *a;
  *a = *b;
  *b = t;
}
int main(void){
  char  a = '1';
  char b = '2';

  swap(&a,&b);
  printf("The value is %c and %c respectively\n",a,b);
  return 0;
}

in the above code, there's a spot that confuse me

I think if a is a pointer, and *a is the value it points to

int *ptr, a = 1;

ptr = &a;
printf("The value of *ptr should be a: %d\n",*ptr);
printf("The value of *a should be an hex address: %p\n",ptr);

so in the swap(char *a, char *b) function,it takes the value not pointer( *a not a),

swap(&a, &b) but it actually pass the pointer value to it as the parameter, and the code works. Anybody can explain it to me?(I think for swap(char *a){...} part, the declaration doesn't mean it require *a to pass in, it means declare a pointer value a, not the value a points to as *ain elsewhere means).

Upvotes: 0

Views: 109

Answers (5)

dknijnal
dknijnal

Reputation: 1

In this our aim is to swap the value

when you write char a='1'; i.e you are putting '1' at address 662442(suppose &a=662442). char b='2'; i.e you are putting '2' at address 662342(suppose &b=662342).

now consider swapping a and b,here our aim is to change value at 662442(&a) to value at 662342(&b) and value at 662342(&b) to value at 662442(&a).

now for swapping we take a temporary variable char temp and will do the following

temp=*a; i.e assigning value at address 662442(&a) to temp.

*a = *b; i.e assigning value at address 662342(&b) to value at address 662442(&a).

now we will put previous value of a (i.e value kept in variable temp) at address 662342(&b)

*b = temp; i.e assigning temp to address 662442(&a)

Upvotes: 0

pb2q
pb2q

Reputation: 59617

Yes, the asterisk * has multiple meanings related to pointers:

  • It be used in a declaration to introduce a variable that contains an address - a pointer:

    int *ptr = NULL;

    Similarly in an argument list, such as void swap(char *, char *).

  • It can also be used to dereference an existing pointer, or get the value pointed to, as within the function swap:

    t = *a;

This generally causes a good deal of confusion for students who are new to C, but it's actually quite simple.

Upvotes: 0

Russell Borogove
Russell Borogove

Reputation: 19037

* is confusing because it means two different, but closely related, things. In a variable declaration, * means "pointer". In an expression, * means "dereference the pointer".

It's intended to be a helpful mnemonic: if you have char *a in your code it means that *a is a char.

Upvotes: 2

Eric J.
Eric J.

Reputation: 150108

Your function

swap(char *a, char *b)

takes two parameters, both of which are of type char *. Quite literally that means they point to a character somewhere in memory.

When you dereference the pointer 

t = *a;

You are saying "grab whatever a is pointing to and put it in t.

Perhaps the confusion is from the fact that * means two related but different things. In the case char *, it's defining a type, specifically one that points to a character somewhere in memory. In the case *a, the * means "look at the character being pointed to by a and let me know what it is".

Upvotes: 2

Ray Toal
Ray Toal

Reputation: 88378

In main:

  • a is a char
  • &a is a pointer to the char a.

In swap:

  • a is a pointer to a char
  • *a is the char pointed to by a.

You passed a pointer to a pointer, and it worked beautifully. Try drawing a picture with boxes and errors. Clears things up every time. Just remember your code has two as and two bs.

Upvotes: 0

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