Ç字符指针问题 [英] c char pointer problem

问题描述

如果我们声明的char * p =你好; 则因为它是写在数据部分中，我们不能修改其中p指向的内容，但我们可以通过修改指针本身。但我发现在C陷阱与缺陷这个例子
安德鲁·柯尼希
AT＆amp; T公司贝尔实验室
美利山，新泽西州07974

if we declare char * p="hello"; then since it is written in data section we cannot modify the contents to which p points but we can modify the pointer itself. but i found this example in C Traps and Pitfalls Andrew Koenig AT&T Bell Laboratories Murray Hill, New Jersey 07974

的例子是

char *p, *q;
p = "xyz";
q = p;
q[1] = ’Y’;

①将指向包含字符串XYZ存储器。所以将P，因为p和q指向相同的存储器

q would point to memory containing the string xYz. So would p, because p and q point to the same memory.

怎么是真的，如果我提到的第一条语句也是如此..
同样地，我跑到下面code

how is it true if the first statement i mentioned is also true.. similarly i ran the following code

main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}

和得到的输出为
兴业银行！gsjfoet

请解释我们如何在这两种情况下，都能够修改内容是什么？
在此先感谢

please explain how in both these cases we are able to modify contents? thanks in advance

推荐答案

您同样的例子使我的系统上分段错误。

Your same example causes a segmentation fault on my system.

您正在运行到未定义的行为在这里。 。数据（注意，字符串可能会在的.text 太）不一定是一成不变的 - 难保机器将写保护内存（通过页表），根据不同的操作系统和编译器。

You're running into undefined behavior here. .data (note that the string literal might be in .text too) is not necessarily immutable - there is no guarantee that the machine will write protect that memory (via page tables), depending on the operating system and compiler.

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