字符串：指针与字符数组 [英] String literals: pointer vs. char array

问题描述

在此语句：

 的char * A =字符串1
 

到底什么是字符串？它是字符串1 ？因为在C 这个线程字符串字面说不同的东西。

到了我的知识

  INT的main（）
{
    字符* A =字符串1; //为只读段分配的内存与字符串文字。
    焦B〔] =字符串2; //就是内存将在堆栈中分配的字符数组。    一个[0] =X; //不允许。这是一个未定义行为。对我来说，赛格故障。
    B [0] ='Y'; //有效。    返回0;
}
 

请添加上述以外的一些细节提到的观点。谢谢你。

调试输出显示错误 A [0] ='Y';

 从/home/jay/Desktop/MI/chararr/a.out...done阅读的符号。
（GDB）b主
在0x40056c断点1：文件ddd.c，4号线。
（GDB）R
启动程序：/home/jay/Desktop/MI/chararr/a.out断点1，主要的（）在ddd.c：4
4 {
（GDB）N
6字符* A =字符串1;
（GDB）N
7 CHAR B〔] =字符串2;
（GDB）
9 [0] =Y;
（GDB）计划接收信号SIGSEGV，分割过错。
0x0000000000400595在main（）在ddd.c：9
 

解决方案

您可以看一下字符串作为的用双引号包围的字符序列的。
该字符串的保存在只读存储器并试图修改此内存导致的未定义行为

那么怎么来的，你得到段错误？的结果
- 主要的一点是，的char * PTR =字符串，使 PTR 指向只读存储器在您的字符串存储。所以，当您尝试访问该存储器： PTR [0] ='X'（这是由方式等同于 *（PTR + 0） ='X'），它是一个内存访问冲突

在另一方面：字符B〔] =字符串2; 分配内存和拷贝字符串字符串2进去，从而改变它是有效的。当 B 超出范围，此内存被释放。

看一看文字字符串初始化一个字符数组

In this statement:

char *a = "string1"

What exactly is string literal? Is it string1? Because this thread String literals in C says something different.

Up to my knowledge

int main()
{
    char *a = "string1"; //is a string- literals allocated memory in read-only section.
    char b[] = "string2"; //is a array char where memory will be allocated in stack.

    a[0] = 'X'; //Not allowed. It is an undefined Behaviour. For me, it Seg Faults. 
    b[0] = 'Y'; //Valid. 

    return 0;
} 

Please add some details other than above mentioned points. Thanks.

Debug Output Showing error in a[0] = 'Y';

Reading symbols from /home/jay/Desktop/MI/chararr/a.out...done.
(gdb) b main
Breakpoint 1 at 0x40056c: file ddd.c, line 4.
(gdb) r
Starting program: /home/jay/Desktop/MI/chararr/a.out 

Breakpoint 1, main () at ddd.c:4
4   {
(gdb) n
6   char *a = "string1";
(gdb) n
7   char b[] = "string2";
(gdb) 
9   a[0] = 'Y';
(gdb) 

Program received signal SIGSEGV, Segmentation fault.
0x0000000000400595 in main () at ddd.c:9

解决方案

You can look at string literal as "a sequence of characters surrounded by double quotes". This string is stored at read-only memory and trying to modify this memory leads to undefined behavior.

So how come that you get segmentation fault?
- The main point is that char *ptr = "string literal" makes ptr to point to the read-only memory where your string literal is stored. So when you try to access this memory: ptr[0] = 'X' (which is by the way equivalent to *(ptr + 0) = 'X'), it is a memory access violation.

On the other hand: char b[] = "string2"; allocates memory and copies string "string2" into it, thus modifying it is valid. This memory is freed when b goes out of scope.

Have a look at Literal string initializer for a character array

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