为什么要在函数标头中声明C数组参数的大小? [英] Why should I declare a C array parameter's size in a function header?

问题描述

有人能启发我为什么要麻烦在函数头中指定C数组参数的大小吗?例如:

Can anyone enlighten me as to why I should bother to specify the size of a C array argument in a function header? For example:

void foo (int iz[6]) { iz[42] = 43; }

使用:

int is[2] = {1,2,3};

我们收到一个有用的错误.也许对评论/文档有帮助吗?

we get a useful error. Perhaps it helps with commenting/documentation?

推荐答案

有人能启发我为什么要麻烦在函数头中指定C数组参数的大小吗?例如:

Can anyone enlighten me as to why I should bother to specify the size of a C array argument in a function header? For example:

void foo(const char sz [6]){sz [42] = 43; }

void foo (const char sz[6]) { sz[42] = 43; }

海事组织，你不应该.当您尝试将数组传递给函数时，真正传递的是指向数组开头的指针.由于该函数接收到的将是一个指针，因此最好编写它使其明确:

IMO, you shouldn't. When you try to pass an array to a function, what's really passed is a pointer to the beginning of the array. Since what the function receives will be a pointer, it's better to write it to make that explicit:

void foo(char const *sz)

然后，由于现在很清楚该函数没有大小提示，因此请将其添加为单独的参数:

Then, since it's now clear that the function has been given no clue of the size, add that as a separate parameter:

void foo(char const *sz, size_t size)

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