为什么c说_Imaginary_I未声明？ [英] Why does clang say _Imaginary_I is not declared?

问题描述

运行

clang test.c -o test

在此文件上

#include <stdio.h>
#include <complex.h>

int main()
{
    _Complex double z = 1.0 + _Imaginary_I * 2.0;
    return 0;
}

由于编译失败

error: use of undeclared identifier '_Imaginary_I'.

根据在线发布，定义了 _Imaginary_I 。

推荐答案

虚数和 _Imaginary_I 是可选功能在C标准中。

Imaginary numbers, and _Imaginary_I, are optional features in the C Standard.

从C11开始，复数也是可选功能，但实现通常支持。 I 和 _Complex_I 应该可以代替。

Complex numbers are also an optional feature as of C11, but are commonly supported by implementations. I and _Complex_I should work instead.

根据标准，您应该能够在编译时通过检查以下宏的值来测试一致性：

According to the standard, you should be able to test for conformance at compile-time by checking the values of the following macros:

• __ STDC_IEC_559_COMPLEX __ ：值 1 表示存在复杂和虚构的类型并符合IEC 60559。


• __ STDC_NO_COMPLEX __ ：值 1 表示既不存在复杂类型也不存在虚类型。

• __STDC_IEC_559_COMPLEX__: value 1 means that complex and imaginary types exist and comply with IEC 60559.
• __STDC_NO_COMPLEX__: value 1 means that neither complex nor imaginary types exist.

但是实际上这是不可靠的，例如 gcc定义了不支持该功能的宏。

However in practice this is not reliable, e.g. gcc defines the macro without supporting the feature.

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