C:将int转换为size_t [英] C: cast int to size_t

问题描述

在32位和64位linux平台上，在C99中将 int 转换/转换为 size_t 的正确方法是什么?

What is the proper way to convert/cast an int to a size_t in C99 on both 32bit and 64bit linux platforms?

示例:

int hash(void * key) {
    //...
}

int main (int argc, char * argv[]) {
    size_t size = 10;
    void * items[size];
    //...
    void * key = ...;
    // Is this the right way to convert the returned int from the hash function
    // to a size_t?
    size_t key_index = (size_t)hash(key) % size;
    void * item = items[key_index];
}

推荐答案

所有算术类型都在C中进行隐式转换.很少需要强制转换-通常仅在要向下转换时才使用，即减少模数1加上最大值.较小的类型，或者当您需要将算术强制为无符号模式以使用无符号算术的属性时.

All arithmetic types convert implicitly in C. It's very rare that you need a cast - usually only when you want to convert down, reducing modulo 1 plus the max value of the smaller type, or when you need to force arithmetic into unsigned mode to use the properties of unsigned arithmetic.

我个人不喜欢看演员表，因为:

Personally, I dislike seeing casts because:

1. 它们看起来很凌乱，
2. 他们向我建议，编写代码的人正在收到有关类型的警告，并在不了解警告原因的情况下进行了强制转换以关闭编译器.

当然，如果您启用了一些超挑剔的警告级别，即使您的隐式转换正确无误，也可能会导致很多警告...

Of course if you enable some ultra-picky warning levels, your implicit conversions might cause lots of warnings even when they're correct...

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