的C指针运算上的字符 [英] C Pointer Arithmetic on Characters

问题描述

我有以下的code，在该行崩溃的我在哪里初始化 CH ：

 的char * p =测试;
    焦炭CH = * P ++;
    的printf（位置：％S \\ N％C，P，CH）;
 

但是以下code没有问题：

 的char * p =测试;
    焦炭CH = * P ++;
    CH ++;
    的printf（位置：％S \\ N％C，P，CH）;
 

解决方案

在第一种情况下，你要修改 T 中的测试字符串编译入程序，这是在你的code未意味着改变存储器的一部分保持（通常;有在那里它允许某些环境中，但通常它是不）。这是因为（* P）++ 手段的（严格意义上）的 * p = * P + 1 （例如，获得字符由 p ，增加它，并把它写回），当然， * p 所指向的编译的测试。

你的第二个版本不存在这样的问题，因为你递增 CH ，您的是的允许改变。你的第二个版本实际上增加两个不同的东西，其实，首先它字符CH = * P ++; 这在检索字符* P 键，然后递增<​​code > p （现在它指向测试中的E），然后你做 CH = CH ++ 。 （你可能意味着只是 CH ++; 在那里，因为 ++ 直接在其操作数时）

I have the following code that crashes at the line where I am initializing ch:

    char * p = "Test";
    char ch = *p++;
    printf("Here : %s\n%c", p, ch);

However the following code has no problem:

    char * p = "Test";
    char ch = *p++;
    ch++;
    printf("Here : %s\n%c", p, ch);

解决方案

In the first situation, you're trying to change the T in the "Test" string compiled into the program, which is held in a part of memory that your code isn't meant to change (usually; there are some environments where it's allowed, but usually it isn't). That's because (*p)++ means (loosely speaking) *p = *p + 1 (e.g., get the character pointed to by p, increment it, and write it back), and of course, *p is pointing to the compiled-in "Test".

Your second version doesn't have that problem, because you're incrementing ch, which you are allowed to change. Your second version actually increments two different things, in fact; first it does char ch = *p++; which retrieves the character at *p and then increments p (now it points to the "e" in "Test"), and then you do ch = ch++. (You probably meant just ch++; there, since ++ operates directly on its operand.)

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