为什么以下会产生分段错误？

Question

int main()
{
        char *temp = "Paras";

        int i;
        i=0;

        temp[3]='F';

        for (i =0 ; i < 5 ; i++ )
                printf("%c\n", temp[i]);

        return 0;
}

Why temp[3]='F'; will cause segmentation fault since temp is not const ?

Answer 1

您不能修改字符串文字。

Answer 2

*temp is defined as a pointer to a constant (sometimes called a string literal - especially in other languages).

Therefore the line with the error is trying to change the third character of this constant.

Try defining a char array and using strcpy to copy temp into it. Then do the above code on the array, it should work. (sorry my ipad here doesn't like to insert code into SO's interface)

Answer 3

As you can see, temp is a pointer, which points to a random address where the nameless array with the value Paras resides. And that array is a string constant.

For your program to work, you need to use an array instead of a pointer:

char temp[6] = "Paras";

Now if you're wondering why it's temp[6] instead of temp[5] , the above code initializes a string, and completely different from:

char temp[5] = {'P', 'a', 'r', 'a', 's'};

Strings are terminated with a null terminator \\0 . And the string initialization will be like:

char temp[6] = {'P', 'a', 'r', 'a', 's', '\0'};

Answer 4

  temp[3]='F'; 

这行不正确。“temp”是const值，所以你不能修改它。