当我在 malloc 内部取消引用一个 NULL 指针时，为什么我的程序没有段错误? [英] Why doesn't my program seg fault when I dereference a NULL pointer inside of malloc?

问题描述

我一直使用这种 malloc 风格

I use this malloc style all the time

int *rc = 0;
rc = malloc(sizeof(*rc));

但是，即使当我调用 sizeof(*rc) 我假设 rc==0 并且我正在取消引用 NULL 指针.

However, it doesn't seg fault even though when I call sizeof(*rc) I assume that rc==0, and I am dereferencing a NULL pointer.

推荐答案

您并没有真正取消引用任何东西.sizeof 的参数不被计算，除非它是一个 VLA.语言明确允许将您想要的任何垃圾"作为 sizeof 的参数.该语言保证它不会评估任何东西，只对表达式的类型进行编译时分析.例如，表达式sizeof i++保证不会改变i的值.

You are not really dereferencing anything. The argument of sizeof is not evaluated, unless it is a VLA. It is explicitly allowed by the language to put whatever "garbage" you want as the argument of sizeof. The language guarantees that it will not evaluate anything, just perform compile-time analysis of the type of the expression. For example, expression sizeof i++ is guaranteed not to change the value of i.

该规则的唯一例外是可变长度数组.对于 VLA，sizeof 的结果是一个运行时值，这意味着该参数已被评估并且必须是有效的.

The only exception from that rule is Variable Length Arrays. The result of sizeof for VLAs is a run-time value, which means that the argument is evaluated and must be valid.

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