在char指针变量中更改char [英] Changing chars in char pointer variable

问题描述

我正在尝试更改char指针变量所指向的chars:

I am trying to change chars pointed to by a char pointer variable:

    char *test3 = "mutable";
    printf("Expected: mutable, Result: %s\n", test3);
    testt(test3);
    printf("Expected tutable, Result: %s\n", test3);

    void testt(char *s) {
        *s = 't'; // FAILS, I get Segmentation Fault Error
    }

为什么上述方法不起作用?指针变量指向的字符是不可变的吗?如果是这样，我将如何修改指针变量的内容?

Why does the above approach not work? Are chars pointed to by pointer variables immutable? If so, how would I modify the contents of the pointer variable?

推荐答案

这是因为您的char *指向字符串文字，并且字符串文字几乎位于只读存储器中的所有现代OS中.

That is because your char * points to a string literal and string literals are in almost every modern OS located in read-only storage.

尝试将其复制到堆栈中:

Try copying it onto the stack:

char test3[] = "mutable";

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