How to swap two numbers without using temp variabl

问题:

This equation swaps two numbers without a temporary variable, but uses arithmetic operations:

a = (a+b) - (b=a);

How can I do it without arithmetic operations? I was thinking about XOR.

回答1:

a=a+b;
b=a-b;
a=a-b;

This is simple yet effective....

回答2:

Why not use the std libs?

std::swap(a,b);

回答3:

In C this should work:

a = a^b;
b = a^b;
a = a^b;

OR a cooler/geekier looking:

a^=b;
b^=a;
a^=b;

For more details look into this. XOR is a very powerful operation that has many interesting usages cropping up here and there.

回答4:

The best way to swap two numbers without using any temporary storage or arithmetic operations is to load both variables into registers, and then use the registers the other way around!

You can't do that directly from C, but the compiler is probably quite capable of working it out for you (at least, if optimisation is enabled) - if you write simple, obvious code, such as that which KennyTM suggested in his comment.

e.g.

void swap_tmp(unsigned int *p)
{
  unsigned int tmp;

  tmp = p[0];
  p[0] = p[1];
  p[1] = tmp;
}

compiled with gcc 4.3.2 with the -O2 optimisation flag gives:

swap_tmp:
        pushl   %ebp               ;  (prologue)
        movl    %esp, %ebp         ;  (prologue)
        movl    8(%ebp), %eax      ; EAX = p
        movl    (%eax), %ecx       ; ECX = p[0]
        movl    4(%eax), %edx      ; EDX = p[1]
        movl    %ecx, 4(%eax)      ; p[1] = ECX
        movl    %edx, (%eax)       ; p[0] = EDX
        popl    %ebp               ;  (epilogue)
        ret                        ;  (epilogue)

回答5:

I haven't seen this C solution before, but I'm sure someone has thought of it. And perhaps had more posting self-control than I do.

fprintf(fopen("temp.txt", "w"), "%d", a);
a = b;
fscanf(fopen("temp.txt", "r"), "%d", &b);

No extra variables!

It works for me, but depending on the stdio implementation you may have to do something about output buffering.

回答6:

a =((a = a + b) - (b = a - b));

回答7:

Using XOR,

void swap(int &a, int &b)
{
    a = a ^ b;
    b = a ^ b;
    a = a ^ b;
}

One liner with XOR,

void swap(int &a, int &b)
{
    a ^= b ^= a ^= b;
}

These methods appear to be clean, because they don't fail for any test-case, but again since (as in method 2) value of variable is modified twice within the same sequence point, it is said to be having undefined behavior declared by ANSI C.

回答8:

In addition to the above solutions for a case where if one of the value is out of range for a signed integer, the two variables values can be swapped in this way

a = a+b;
b=b-(-a);
a=b-a;
b=-(b);

回答9:

Multiplication and division can also be used.

 int x = 10, y = 5;

 // Code to swap 'x' and 'y'
 x = x * y;  // x now becomes 50
 y = x / y;  // y becomes 10
 x = x / y;  // x becomes 5

回答10:

Simple solution which i remember from my bachelors :-)

a = a+b-(b=a);

回答11:

Just use something else, e.g. a variable that's not a temporary. For instance,

int main (int argc, char** argv) {
   int a = 5; int b = 6;
   argc = a; a = b; b = argc;
}

After all, the point of the question is not to show the sane way to do it (c=a;a=b;b=c). It's to show you can think out of the box, or at least copy the answer of someone else who can.