strcpy() 返回值 [英] strcpy() return value

问题描述

标准 C 库中的许多函数，尤其是用于字符串操作的函数，尤其是 strcpy()，共享以下原型:

A lot of the functions from the standard C library, especially the ones for string manipulation, and most notably strcpy(), share the following prototype:

char *the_function (char *destination, ...)

这些函数的返回值实际上与提供的destination相同.为什么要为多余的东西浪费返回值?将这样的函数设为 void 或返回有用的东西更有意义.

The return value of these functions is in fact the same as the provided destination. Why would you waste the return value for something redundant? It makes more sense for such a function to be void or return something useful.

我唯一的猜测是，将函数调用嵌套在另一个表达式中更容易、更方便，例如:

My only guess as to why this is is that it's easier and more convenient to nest the function call in another expression, for example:

printf("%s
", strcpy(dst, src));

是否有任何其他合理的理由来证明这个习语是正确的?

Are there any other sensible reasons to justify this idiom?

推荐答案

正如 Evan 指出的那样，可以做类似的事情

as Evan pointed out, it is possible to do something like

char* s = strcpy(malloc(10), "test");

例如为 malloc()ed 内存分配一个值，不使用辅助变量.

e.g. assign malloc()ed memory a value, without using helper variable.

(这个例子不是最好的，它会在内存不足的情况下崩溃，但这个想法很明显)

(this example isn't the best one, it will crash on out of memory conditions, but the idea is obvious)

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