一个C code的输出像通过引用调用 [英] Output of a c code like with call by reference

问题描述

鉴于这种C code：

Given this C code:

int x=12,y=10;

void tswap(int pa, int pb) {
   int tmp;
   tmp=pa;
   pa=pb;
   pb=tmp;
   x=x+pa;
   x=x-pb;
   y++;
   printf("%d %d %d %d\n",pa,pb,x,y);
}

int main() {
    int a=4;
    tswap(x,a);
    printf("%d %d %d\n",x,y,a);
    return 0;
}

我要弄清楚什么如果C使用的call-by-参考它会打印。

I have to figure out what it would print if C used call-by-reference.

下面是我的期望。

首先
的
X = 12 Y = 10和= 4

Here is my expectations.
First of all
x=12 y=10 and a=4

after tswap(12,4):
pa=x=12 pb=a=4

after tmp=pa:
tmp=pa=x=12
-----------
after pa=pb:
tmp=x=12
pa=pb=a=4
-------------
after pb=tmp:
pb=tmp=x=12
pa=a=4

after x=x+pa
x=16
------------
and after x=x-pb and y++
x=4,y=11

所以我认为X = 4，PB = 12，PA = 4，Y = 11;但得到的答复说-4 -4 12 11.如果我错了？

So I think x=4, pb=12, pa=4, y=11; but the answer says -4 12 -4 11. Where I am wrong?

推荐答案

由于没有递归在这里，所有的变量都具有不同的名称，我们可以骗了一下，改变局部变量，全局（因为这不会造成任何名称冲突）：

Since there's no recursion here, and all variables have distinct names, we can "cheat" a bit and change the local variables to globals (since this won't cause any name conflicts):

int x=12, y=10; /* original globals */
int tmp;        /* belongs to tswap */
int a;          /* belongs to main */

void tswap(int pa, int pb) {
   tmp=pa;
   pa=pb;
   pb=tmp;
   x=x+pa;
   x=x-pb;
   y++;
   printf("%d %d %d %d\n",pa,pb,x,y);
}

int main() {
    a=4;
    tswap(x,a);
    printf("%d %d %d\n",x,y,a);
    return 0;
}

由于 tswap 只调用一次，我们知道它的每年参数的总是别名为 X ，其 PB 参数的总是的别名为 A 。因此，我们就可以摆脱的参数，使用变量替换他们，他们的别名：

Since tswap is only called once, we know that its pa parameter is always aliased to x, and its pb parameter is always aliased to a. So, we can just get rid of the parameters, replacing them with the variables that they alias:

int x=12, y=10;
int tmp;
int a;

void tswap() {
   tmp=x;
   x=a;
   a=tmp;
   x=x+x;
   x=x-a;
   y++;
   printf("%d %d %d %d\n",x,a,x,y);
}

int main() {
    a=4;
    tswap();
    printf("%d %d %d\n",x,y,a);
    return 0;
}

在这一点上，这只是常规的C code;我们不再有任何参数，因此传递通过引用和传递按值是等价的。通过追踪：

At this point, this is just regular C code; we no longer have any parameters, so "pass-by-reference" and "pass-by-value" are equivalent. Tracing through:

int x=12, y=10;
int tmp;
int a;

void tswap() {
   tmp=x;       /* x=12, y=10, tmp=12, a=4 */
   x=a;         /* x=4, y=10, tmp=12, a=4 */
   a=tmp;       /* x=4, y=10, tmp=12, a=12 */
   x=x+x;       /* x=8, y=10, tmp=12, a=12 */
   x=x-a;       /* x=-4, y=10, tmp=12, a=12 */
   y++;         /* x=-4, y=11, tmp=12, a=12 */
   printf("%d %d %d %d\n",x,a,x,y);    /* prints "-4 12 -4 11" */
}

int main() {
    a=4;        /* x=12, y=10, tmp is uninitialized, a=4 */
    tswap();    /* x=-4, y=11, tmp=12, a=4; prints "-4 12 -4 11" */
    printf("%d %d %d\n",x,y,a); /* prints "-4 11 4" */
    return 0;
}

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