有关指针用C [英] about Pointers in C

问题描述

我开始阅读关于C指针几篇文章和我有一个例子，我不明白。
应该是什么以下code的输出??

I started to read a few articles about pointers in C and I've got one example that I don't understand. What should be the output of following code..??

    main()
     {
      char far *s1 ,*s2;
      printf("%d,%d",sizeof(s1),sizeof(s2));
     }

OUTPUT-4,2

OUTPUT-4,2

据我来说，双方的sizeof返回的值（）函数应该是4，因为一个远指针有4个字节的地址。

According to me, value returned by both sizeof() functions should be 4 because a far pointer has 4 byte address.

但解决方案手册中的答案是4,2。任何一个可以解释？
任何人都可以解释PLZ> ???

but the answer in solution manual is 4,2. Can any one explain ?? can anyone plz explain>???

推荐答案

这是一样的书写

char far *s1;
char *s2;
the "far" is not distributed across all variables, e.g.
char far *s1, ch;

据使得在一个普通的字符ch没有意义的。

far makes no sense on a normal character ch.

因此​​S2不是远的指针，并且作为接近指针，这是16位宽在您的目标进行处理。

Hence s2 is not a "far" pointer, and is handled as a "near" pointer, which is 16 bits wide in your target.

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