printf语句间$ P $如何PTED？ [英] How is printf statement interpreted?

问题描述

如何以下行间$ P $由GCC编译器PTED：

How is the following line interpreted by GCC compiler:

printf("HELLO");  

我想知道这是因为当我运行下面的程序：

I want to know this because when I am running following program:

main()  
{  
    printf(5+"Good Morning");  
}  

该计划是打印：

Morning

为什么编译器会从第6个字符开始打印？

Why is the compiler is starting the printing from the sixth character?

推荐答案

有这里发生了很多事情。正如其他人所说，的printf（）不'知道'对前pression 5+早上好什么。即前pression的价值是由C语言决定。

There are a lot of things happening here. As others have said, printf() doesn't 'know' anything about the expression 5+"Good Morning". The value of that expression is determined by the C language.

首先， A + B 相同 B + A ，所以 5 +早上好相同早上好+5 。

First, a+b is the same as b+a, so 5+"Good Morning" is the same as "Good Morning"+5.

现在，早上好的类型（即一个字符串）是一个阵列字符。具体来说，早上好是一个13字符数组（12常规字，随后 0 ） 。当大多数前pressions使用时，输入C数组衰变的指针到它的第一要素，二进制加法就是这样的一个情况。这一切都意味着，在早上好+5 ，早上好衰变到一个指向它的第一个元素，这是字符。

Now, the type of "Good Morning" (i.e., a string literal) is an "array of char". Specifically, "Good Morning" is a 13-character array (12 "regular" characters, followed by a 0). When used in most expressions, the type of an array in C "decays" to a pointer to its first element, and binary addition is one such case. All this means that in "Good Morning"+5, "Good Morning" decays to a pointer to its first element, which is the character G.

下面是内存的样子：

  0   1   2   3   4   5   6   7   8   9   0   1   2
+---+---+---+---+---+---+---+---+---+---+---+---+---+
| G | o | o | d |   | M | o | r | n | i | n | g | 0 |
+---+---+---+---+---+---+---+---+---+---+---+---+---+

的地址加上5的值是指向从上面的5个位置的指针，这是 M 。因此，的printf（）越来越即在 M 的地址。即，直到它找到一个 0 的printf（）打印。因此，你看到晨作为输出。

The value of the address of G plus 5 is a pointer that points to 5 locations from G above, which is M. So, printf() is getting an address that is at M. printf() prints that till it finds a 0. Hence you see Morning as output.

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