我可以用“字节数"调用 memcpy() 和 memmove() 吗?设置为零?

问题描述

当我实际上没有什么可移动/复制的情况时，我是否需要将 memmove()/memcpy() 作为边缘情况处理

Do I need to treat cases when I actully have nothing to move/copy with memmove()/memcpy() as edge cases

int numberOfBytes = ...
if( numberOfBytes != 0 ) {
    memmove( dest, source, numberOfBytes );
}

或者我应该直接调用函数而不检查

or should I just call the function without checking

int numberOfBytes = ...
memmove( dest, source, numberOfBytes );

是否需要检查之前的代码片段?

Is the check in the former snippet necessary?

推荐答案

来自 C99 标准 (7.21.1/2):

From the C99 standard (7.21.1/2):

声明为 size_t n 的参数指定数组的长度函数，n 在调用该函数时可以具有零值.除非明确说明否则在本小节中对特定函数的描述中，指针参数此类调用仍应具有有效值，如 7.1.4 中所述.在这样的电话中，一个定位一个字符的函数没有找到，一个比较两个字符的函数字符序列返回零，复制字符的函数复制零字符.

Where an argument declared as size_t n specifies the length of the array for a function, n can have the value zero on a call to that function. Unless explicitly stated otherwise in the description of a particular function in this subclause, pointer arguments on such a call shall still have valid values, as described in 7.1.4. On such a call, a function that locates a character finds no occurrence, a function that compares two character sequences returns zero, and a function that copies characters copies zero characters.

所以答案是否定的；不需要检查(或者是；您可以通过零).

So the answer is no; the check is not necessary (or yes; you can pass zero).

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