我们可以创建一个IF语句来确定变量是否已在C / C ++中声明为指针 [英] Can we create an IF statement that determines if a variable has been declared as a pointer in C/C++

问题描述

假设变量 pa 总是被声明为以下两种方式之一：

Suppose a variable pa is always declared as one of two ways:

double * pa

double * pa

OR

double pa

double pa

我们可以创建一个执行以下操作的IF语句

Can we create an IF statement that does the following

IF（pa是一个指针）
{pa [0] = 1
}

IF (pa is a pointer) { pa[0] = 1 }

ELSE
{pa = 1}

ELSE { pa = 1}

编辑：

请参阅如何确定指针数组是否有已填写C ++ / C 后续问题

推荐答案

不是直接，不是。假设有一种方法可以做到这一点，例如

Not directly, no. Suppose there were a way to do this, something like

if (__IS_POINTER(pa) {
    pa[0] = 1;
}
else {
    a = 1;
}

如果pa 不是指针会怎么样？你的程序中仍然有 pa [0] = 1; 语句，这是非法的，所以编译器会拒绝你的程序。

What happens if pa isn't a pointer? You still have that pa[0] = 1; statement in your program, and it's illegal, so the compiler is going to reject your program.

你原则上可以进行编译时测试：

You might in principle be able to do a compile-time test:

#if __pa_IS_A_POINTER
pa[0] = 1;
#else
a = 1
#endif

但是C ++预处理器的功能非常有限;它无法测试a的类型变量或表达。

but the power of the C++ preprocessor is very limited; it has no way to test the type of a variable or expression.

如果你有这个能力，你会用它做什么？告诉我们你的实际目标是什么。

If you had this capability, what would you do with it? Tell us what your actual goal is.

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