#include <stdio.h> 
 
void main(){ 
  int a = 3; 
  int b = 4; 
 
  a = a ^ b;//使用異或交換 
  b = b ^ a; 
  a = a ^ b; 
 
  printf("%d, %d\n", a, b); 
 
  a = a - b;//使用加減交換 
  b = a + b; 
  a = b - a; 
 
  printf("%d, %d\n", a, b); 
 
  a ^= b ^= a ^= b; 
 
  printf("%d, %d\n", a, b); 
} 

#include <stdio.h> 
#include <stdlib.h> 
 
int sumof1(int x)//求一個(gè)數(shù)轉(zhuǎn)換成二進(jìn)制以后1的個(gè)數(shù) 
{ 
  int countx = 0; 
  while(x) 
  { 
    countx ++; 
    x &= x-1; //每位與一次x - 1；就能消掉最后一個(gè)1 
  } 
  return countx; 
} 
 
void main(){ 
 
  char c[10]; 
  int i = 999; 
 
  itoa(i, c, 10);//以10進(jìn)制轉(zhuǎn)換成字符數(shù)組 
  puts(c); 
 
  itoa(i, c, 16);//以16進(jìn)制轉(zhuǎn)換成字符數(shù)組 
  printf("0x%s\n", c); 
 
  itoa(i, c, 8);//以8進(jìn)制轉(zhuǎn)換成字符數(shù)組 
  printf("0%s\n", c); 
 
  itoa(i, c, 2);//以2進(jìn)制轉(zhuǎn)換成字符數(shù)組 
  puts(c); 
 
  i = atoi(c);//再將字符串轉(zhuǎn)成整形 
  printf("%d\n", i); 
 
  printf("%d\n", sumof1(i));//以2進(jìn)制表示時(shí)1的個(gè)數(shù) 
}