返回值是多少？

发布于 2024-12-14 07:07:16 字数 211 浏览 2 评论 0原文

在通过引用传递参数的语言中，给定以下函数：

int function g(x, y) {
  x = x + 1;
  y = y + 2;
  return x + y;
}

如果调用i = 3，并且调用g(i,i)，则返回值是多少？我以为是9，这是正确的吗？

原文

In a language that passes parameters by reference, given the following function:

int function g(x, y) {
  x = x + 1;
  y = y + 2;
  return x + y;
}

If i = 3, and g(i,i) is called, what is value returned? I thought it is 9, is this correct?

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舂唻埖巳落 2024-12-21 07:07:16

如果它是按引用传递（你原来的问题是 C，但 C 没有按引用传递，而且问题从那时起就发生了变化，所以我会笼统地回答），情况可能是 x 和 y 将简单地修改为它们传入的变量。毕竟，这就是参考。

在这种情况下，它们都是对相同变量i的引用，所以你的序列可能是：

i = i + 1;       // i becomes 4.
i = i + 2;       // i becomes 6.
return i + i;    // return i + i, or 12.

你可以在使用以下 C 进行操作（使用指针模拟传递引用）：

pax$ cat qq.c
#include <stdio.h>

int g(int *x, int *y) {
    *x = *x + 1;
    *y = *y + 2;
    return *x + *y;
}

int main (void) {
    int i = 3;
    int rv = g (&i, &i);
    printf ("Returned: %d\n", rv);
    return 0;
}

pax$ gcc -o qq qq.c ; ./qq
Returned: 12

您的结果 9 似乎假设引用彼此不同，例如在以下代码中：（

#include <stdio.h>

int g(int *x, int *y) {
    *x = *x + 1;
    *y = *y + 2;
    return *x + *y;
}

int main (void) {
    int i1 = 3, i2 = 3;
    int rv = g (&i1, &i2);
    printf ("Returned: %d\n", rv);
    return 0;
}

这确实输出 9) 但通常情况并非如此参考类型。

If it's pass-by-reference (your original question was C but C doesn't have pass-by-reference and the question has changed since then anyway, so I'll answer generically), it's probably the case that x and y will simply modify the variables that are passed in for them. That's what a reference is, after all.

In this case, they're both a reference to the same variable i, so your sequence is likely to be:

i = i + 1;       // i becomes 4.
i = i + 2;       // i becomes 6.
return i + i;    // return i + i, or 12.

You can see this in operation with the following C (using pointers to emulate pass-by-reference):

pax$ cat qq.c
#include <stdio.h>

int g(int *x, int *y) {
    *x = *x + 1;
    *y = *y + 2;
    return *x + *y;
}

int main (void) {
    int i = 3;
    int rv = g (&i, &i);
    printf ("Returned: %d\n", rv);
    return 0;
}

pax$ gcc -o qq qq.c ; ./qq
Returned: 12

Your result of 9 seems to be assuming that the references are distinct from one another, such as in the following code:

#include <stdio.h>

int g(int *x, int *y) {
    *x = *x + 1;
    *y = *y + 2;
    return *x + *y;
}

int main (void) {
    int i1 = 3, i2 = 3;
    int rv = g (&i1, &i2);
    printf ("Returned: %d\n", rv);
    return 0;
}

(this does output 9) but that's not usually the case with reference types.

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