指针条件 while(*s1++=*s2++) [英] Pointer Condition while(*s1++=*s2++)

问题描述

int main()
{
char str1[] = "Overflow";
char str2[] = "Stack";
char *s1 = str1, *s2=str2;
while(*s1++ = *s2++)
printf("%s", str1);
return 0;
}

当这个条件打破时

(while(*s1++ = *s2++))

请解释一下逻辑

输出是

推荐答案

*s1++，在这个变量中，*的优先级高于++.在下面的代码中，

*s1++, in this variable, * is having higher precedence than ++. In below code,

while(*s1++ = *s2++); //There should be a semicolon.Your code does not has.

1. *s1 = *s2 首先评估.
2. C 中几乎每个表达式都返回一个值.所以 *s1 = *s2 返回一个值，它是第一个字符，显然不是空的.while() 循环被执行.
3. 然后 s1 和 s2 递增.
4. 当s2到达字符串末尾时，*s2返回'\0'，分配给s1>.*s1=*s2.现在这个表达式也返回一个值，它是 '\0' 并且 while('\0') 循环终止.

1. *s1 = *s2 is evaluated first.
2. Almost every expression in C returns a value. So *s1 = *s2 returns a value which is the first character and its obviously not null. while() loop is executed.
3. Then s1 and s2 are incremented.
4. When s2 reaches the end of string, *s2 returns '\0' which is assigned to s1. *s1=*s2. Now this expression also return a value which is '\0' and while('\0') loop terminates.

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