不使用其他變量交換兩個(gè)整型的值：
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#include <stdio.h> void main(){ 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); } |
整形和字符數(shù)組型轉(zhuǎn)換：
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#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) } |
