指向字符串的指针 [英] pointer to strings

问题描述

下面的两个代码段可编译，没有任何错误/警告，但运行时会崩溃.请赐教.

The following two code snippets compiles without any errors/warnings but while running it crashes. Kindly enlighten me.

程序1

 int main( )
{
   char *p= "Hello" ;

   *p = 'B' ;
    printf("\n%s",p);

   return 0;
}

程序2

int main( )
{
   char *p= "Hello" ;
   Char *q="mug"
   *q = *p ;
    printf("\n%s",q);

   return 0;
}

对于程序2，我希望输出为拥抱".

For program 2 i expected output to be 'Hug'.

推荐答案

当您这样做:

char *p= "Hello";

您正在定义字符串文字.字符串文字是常量数据，正如您所发现的那样，对其进行修改会导致未定义的行为(通常是崩溃).应该声明为:

You are defining a string literal. String literals are constant data and as you've found out, modifying them results in undefined behavior (often a crash). It should be declared as:

const char *p = "Hello";

因此，如果您尝试修改它，编译器将抛出错误.

So the compiler will throw an error if you try to modify it.

现在，如果您将其定义为:

Now if you define it instead as:

char p[] = "Hello";

然后将内存分配到堆栈上，您可以对其进行修改.

The memory is then allocated on the stack and you can modify it.

int main(int argc, char *argv[])
{
    char p[] = "Hello" ;

    *p = 'B' ;
    printf("\n%s",p);

    return 0;
}

输出Bello

对于程序2，请注意只需将q放在堆栈中. p可以仍然是指向字符串文字的const指针，因为您只是从中读取.

For program 2, note only q needs to be on the stack. p can remain a const pointer to a string literal, since you're only reading from it.

int main( )
{
    const char *p = "Hello" ;
    char q[] = "mug";
    *q = *p ;
    printf("\n%s",q);

    return 0;
}

输出Hug

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