有没有一种方法可以在不创建变量的情况下调用需要指针的函数? [英] Is there a way to call a function expecting a pointer without creating variable?

问题描述

我有这个函数调用:

uint32_t func(uint32_t* a, uint32_t b)

我想用这样的整数字面量来调用它:

I want to call it with an integer literal like this:

func(0, b);

其中b是uint32_t.

有什么办法可以在不创建中间变量的情况下做到这一点?
IE.我想避免这样做:

Is there any way I can do this without creating an intermediate variable?
I.e. I want to avoid doing this:

uint32_t a = 0;
func(a, b);

推荐答案

帮助程序类:

struct int_ptr {
    int v;

    operator int *() { return &v; }
};

int foo(int *a, int b);

void bar()
{
    foo(int_ptr{0}, 0);
}

这将导致构造一个临时int_ptr类，将其v成员初始化为0.这将作为参数传递给采用int *的函数，并且int_ptr提供合适的将右指针传递给函数的方法.

This results in a construction of a temporary int_ptr class, initializing its v member to 0. This gets passed as a parameter to a function that takes an int *, and int_ptr provides a suitable operator * method that passes the right pointer to the function.

这整个牌库取决于int_ptr临时目录存在直到函数调用结束的事实.您应该为助手类选择一个名称来强调这一事实.如果您始终使用它将指向0的指针传递给foo，则将其拼写出来:

This entire house of cards hinges on the fact that the int_ptr temporary exists until the end of the function call. You should pick a name for the helper class to underline that fact. If you always use it to pass a pointer to 0 to foo, then spell it out:

struct zero_value_to_foo {
    int v=0;

    operator int *() { return &v; }
};

int foo(int *a, int b);

void bar()
{

    foo(zero_value_to_foo{}, 0);
}

因此在其他情况下使用它看起来很不合适，即

So that using it in other contexts will look to be very much out of place, i.e.

int *p=zero_value_to_foo{};

这可以编译，但是留下一个悬空的指针；但是希望"zero_value_to_foo"标签可以提供一个暗示性的提示，说明这里有问题.

This compiles, but leaves you with a dangling pointer; but hopefully the "zero_value_to_foo" label gives a honking clue that something is wrong here.

要避免误用，您可以做的另一件事是为操作员使用ref限定符:

Another little thing you can do to help yourself from misusing this is to use a ref qualifier for the operator:

struct zero_value_to_foo {
    int v=0;

    operator int *() && { return &v; }
};

有了这个，

foo(zero_value_to_foo{}, 0);

仍然可以编译，但不能编译:

still compiles, but not this:

zero_value_to_foo zero{};

foo(zero, 0);

除了在上下文中意味着这样做以外，可以做更多的工作来使其难以使用.

The more that can be done to make it difficult to use this except in the context is meant for, the fewer opportunities there are for bugs to creep by.

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