First of all, start explaining from mounting by the most basic function.
Naturally, use the argument of the pointer type because it is necessary to change the value of two variable.
Notes here : though it only has to exchange two variables on the side of the function now.

 void swap(int *a,int *b)  
{
    *a = *b;  
    *b = *a;  
    return;
}

 
#include <stdio.h> 

void swap(int *a,int *b);  

int main(void)  
{
    int a = 10,b = 100;  
    printf("a = %3d : b = %3d\n",a,b);     &
nbsp;swap(&a,&b);  
    printf("a = %3d : b = %3d\n",a,b);  
    return 0;  
}

void swap(int *a,int *b)  
{
    int temp;  
    temp = *a;  
    *a = *b;  
    *b = temp;  
    return;
}

a = 10 : b = 100
a = 100 : b = 10

It is mounting by the following addition and subtraction to be thought first of all.

 
#define SWAP(a,b) (a += b,b = a - b,a -= b)

It began to devise the method using another one and the exclusive-OR.

 
#define SWAP(a,b) (a ^= b,b = a ^ b,a ^= b)

Though these two macros operate without trouble in case of almost
Actually, there is a case that doesn't operate well, and it becomes the maximum problem.
It happens when the same variable is specified.
Will actually try from the theory because it is evidence.

 
#include <stdio.h> 

#define SWAP(a,b) (a += b,b = a - b,a -= b)  

int main(void)  
{
    int a = 98;  
    printf("a = %3d\n",a);  
    SWAP(a,a);  
    printf("a = %3d\n",a);  
    return 0;
}

a = 98
a = 0

Though it might not be thought that there is what of the exchange of the same variables
Actually, with the sort program with array[i] When you exchanged array[j] It was i == j.
Then, be and get it. Moreover, most of the usage of SWAP is a sort program.

Generally though it doesn't depress any further because this problem is helpless
The author wants to solve this even forcibly, and is using the method by & operator.
In & operator, there is a character that the following expression is not executed if the expression executed immediately before is an imitation.
The following macro is SWAP macro using the character of & operator.

 #define SWAP(a,b) ((a != b) && (a += b,b = a - b,a -= b))

Or, clause 3 operator can be used so as not to depend on the character of the operator.

 #define SWAP(a,b) ((a != b)? (a += b,b = a - b,a -= b) : 0 )

Though the SWAP macro can be made if the pointer variable is not used
Only the exchange of pointer variables has only mounting by the function shown first by all means.
And, pointers might be exchanged with the sort program unfortunately.

In addition, there is a possibility that drop out happens when the real number values are exchanged.
For instance, when the value of variable a is terribly huge, and the value of b is terribly minimum
The result is rounded even if a+b is calculated, and there is a possibility of becoming it as well as a.
Therefore, when accuracy is requested, it is not possible to use it.

After all, to exchange the values surely, the variable is temporarily needed by all means.
It is necessary to declare a variable in the macro for that.

If the type is specified, it is possible to declare a variable because the macro is only a mere replacement.
Moreover, the collision of the variable identifier can be avoided if it encloses and it blocks it.
Concretely, it becomes the following.

 #define SWAP(type,a,b) { type temp = a; a = b; b = temp; } 

The following program is an example of exchanging variables by using this macro.

 
#include <stdio.h> 

#define SWAP(type,a,b) { type temp = a; a = b; b = temp; } 

int main(void)  
{
	int a = 10,b = 100;  
	printf("a = %3d : b = %3d\n",a,b);  
	SWAP(int,a,b) printf("a = %3d : 
	b = %3d\n",a,b);  
	return 0;
}

a = 10 : b = 100
a = 100 : b = 10

At the end;The person of yucky : [wotsukerarenaino].

 #define swap(type,a,b) do{type _c;_c=a;a=b;b=_c;}while(0)

Because it is not concluded only by the macro as a sentence if it makes it
At the end;[Wotsukete] comes to operate.