“ c＆quot”中功能中的数据操作

发布于 2025-01-29 11:38:20 字数 866 浏览 3 评论 0原文

这是我不明白的事情。也许有人可以阐明它。

我知道数据操作的标准方法正在传递对函数的引用并更改函数中的数据。这样：

#include <stdio.h>
#include <stdlib.h>


void function2( float *param) {
    printf("I've received value %f\n", *param);
    (*param)++;
}

int main(void) {
    float variable = 111;

    function2(&variable);
    printf("variable %f\n", variable);
    return 0;
}

当使用（＆amp; varible）调用功能2时，我希望该函数可以更改数据。到目前为止，还没有问题。

但是为什么这也起作用呢？

#include <stdio.h>
#include <stdlib.h>


void function2( float &param) {
    printf("I've received value %f\n", param);
    (param)++;
}

int main(void) {
    float variable = 111;

    function2(variable);
    printf("variable %f\n", variable);
    return 0;
}

在我的理解中，当调用函数2（变量）时，“变量”值的副本将传递给函数。但是，在函数调用后，“变量”的值已更改。

在阅读这样的代码时，无论功能内部发生什么，我都不会期望“变量”的数据会更改。

原文

Here's something I don't understand. Maybe someone can shed some light on it.

I know that the standard method for data manipulation is passing a reference to a function and changing the data in the function. Like this:

#include <stdio.h>
#include <stdlib.h>


void function2( float *param) {
    printf("I've received value %f\n", *param);
    (*param)++;
}

int main(void) {
    float variable = 111;

    function2(&variable);
    printf("variable %f\n", variable);
    return 0;
}

When calling function2 with (&variable) I expect that the function can change the data.
So far no questions.

But why does this work as well?

#include <stdio.h>
#include <stdlib.h>


void function2( float ¶m) {
    printf("I've received value %f\n", param);
    (param)++;
}

int main(void) {
    float variable = 111;

    function2(variable);
    printf("variable %f\n", variable);
    return 0;
}

In my understanding, when calling function2(variable) a copy of the value of "variable" is passed to the function. But nevertheless the value of "variable" has changed after the function call.

When reading a code like this I would never expect the data of "variable" to change, no matter what happens inside the function.

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晚风撩人 2025-02-05 11:38:20

该参数的声明

void function2( float ¶m) {

是不正确的。在C中没有参考。此类函数声明将在C ++中有效，其中存在引用。

当函数被调用时，就像

function2(variable);

创建了变量的副本一样。该函数是指主声明的原始变量。

This declaration of the parameter

void function2( float ¶m) {

is incorrect. In C there is no reference. Such a function declaration will be valid in C++, where references exist.

In C++ when the function is called like

function2(variable);

neither copy of the variable is created. The function refers to the original variable declared in main.

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任谁 2025-02-05 11:38:20

变化1期望A 指针作为参数。因此，您需要明确传递＆amp; variable。

变体2期望a 参考作为参数。它们像一个指针一样 Kinda ，因为它们仍然指向内存中的原始位置，但也有些不同，因为它们键入明智，比单纯的记忆中的地址要多。就像，如果我没记错的话，您不能使用引用进行时髦的指针算术。

如您所见，如果一个函数期望参考作为参数，则编译器将知道该怎么做，而无需明确创建参考。

另外，如您所见，如果有人调用您的功能，并且没想到它会更改原始数据，这可能会带来令人讨厌的惊喜。这就是为什么在干净的代码中，您不会引入意外的副作用；如果功能与输入混乱，那应该是唯一的事情，并且应该非常明确。

要最终获得一个不做 更改原始的变体，请

void function2( float param)

在函数声明中进行。否*和否＆amp;。在这种情况下，该函数不会获得指针或参考，而只是仅获得浮点值，现在将是输入参数的 copy ，因此不会更改您的变量。

Variation 1 expects a pointer as argument. Hence you need to explicitly pass in &variable.

Variation 2 expects a reference as argument. They're kinda like a pointer in that they still point to the original place in memory, but also a bit different in that they're, type wise, a bit more than a mere address in memory. Like, you can't do funky pointer arithmetic with a reference if I recall correctly.

As you can see, if a function expects a reference as argument, the compiler will know what to do without you having to explicitly create a reference.

Also, as you can see, this has the potential for nasty surprises if someone calls your function and didn't expect it to change the original data. That's why in clean code you don't introduce unexpected side effects; if a function messes with input that should be the only thing it does and it should be quite explicit.

To finally get a variation that does not change the original, do

void function2( float param)

in the function declaration. No * and no &. In this case the function doesn't get a pointer or a reference but simply a float value, which will now be a copy of the input argument and thus won't change your variable.

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