C语言中的指针和变量之间的区别? [英] Differences between pointers and variables in C?

问题描述

#include <stdio.h>
int inc1(int x)  { return x++;    }
int inc2(int *x) { return (*x)++; }

int
main(void)
{
    int a;
    a = 3;
    printf("%d\n", inc1(a) + a);
    printf("%d\n", inc2(a) + a);
    return 0;
}

我正在研究过去的论文，其中一个问题是跟踪第6行和第9行之间对a所做的更改.我有点理解指针(指的是内存位置)，但是如果有人可以通过在这段代码中所做的更改非常棒.

I'm working through a past paper and one of the questions is to track the changes made to a between lines 6 and 9. I kind of understand pointers (referencing to memory location) but if someone could just talk me through the changes made to a throughout this piece of code that would be great.

推荐答案

我将解释这个几乎相同的代码，其中不包含您的帖子中的错误:

I'll explain this nearly identical code that doesn't contain the error in your post:

#include <stdio.h>

int inc1(int x)  { return x++;    }
int inc2(int *x) { return (*x)++; }

int main(void) {
    int a;
    a = 3;
    printf("%d\n", inc1(a) + a);
    printf("%d\n", inc2(&a) + a);
    return 0;
} 

a初始化为3，然后将 a 的值传递给inc1()，该值将其返回并使用 post-increment 加1.这意味着返回的实际值仍为3.

a is initialized to 3, then the value of a is passed to inc1(), which returns it and adds 1 using post-increment. This means the actual value returned is still 3.

接下来， a 的地址被传递到inc2().这意味着x中的值发生了什么变化.同样，使用后增量，所以inc2()返回的是3，但是在调用之后，a是4.

Next, the address of a is passed to inc2(). That means that what happens to the value in x happens to a. Again, post-increment is used, so what inc2() returns is 3, but after the call, a is 4.

但是，编译器可以自由地在序列点之间以任何顺序求值诸如a或inc2(&a)的表达式.这意味着inc2(&a) + a右边的a可以是3或4(取决于a是在int2(&a)之前还是之后进行求值的，因此程序可以输出 6 7 >或 6 6 .

However, compilers are free to evaluate expressions, such as a or inc2(&a), in any order between sequence points. This means that the a on the right of inc2(&a) + a may be either 3 or 4 (depending on whether a is evaluated before or after int2(&a), so the program may output either 6 7 or 6 6.

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