问题描述
在以下代码中
char str [] =" asdf" ;
str [0] ='s'';
是可能的。
但是char * str =" ASDF" ;
str [0] ='s'';
是运行时错误。
我理解的是什么是* str =" asdf"存储在只读存储器中。
as str []存储在堆栈中。
有没有更好的解释呢？
In following code
char str[] = "asdf" ;
str[0] = ''s'' ;
is possible.
But char *str = "asdf" ;
str[0] = ''s'' ;
is an run-time error.
What i understand is *str = "asdf" is stored in Read-only-Memory. Where
as str[] is stored in stack.
Does any body has any better explanation for that?
推荐答案
str [ ]是数据的副本asdf
* str指向asdf本身。
str[] is a copy of the data "asdf"
*str points to "asdf" itself.
str []是副本数据asdf
* str指向asdf本身。
str[] is a copy of the data "asdf"
*str points to "asdf" itself.
除此之外，还有未定义的行为要修改
a string litteral - 这就是他在指针
case。
And in addition to that, it''s undefined behavior to modify
a string litteral - which is what he did in the pointer
case.
它是不允许更新字符串文字，因此如果你这样做，可能会发生任何
。有些机器会崩溃，有些机器会改变字面值，有些机器会覆盖[伪]随机内存，有些会写诽谤性电子邮件
给你当地的国家元首。 />
有些实现有一个堆栈，但是`str []`不需要在它上面。
有些实现有只读内存，但是"不需要在其中。
但是不允许更新字符串文字。
-
克里斯尝试，或尝试不 - 没有做 Dollin
It is Not Allowed to update string literals, so if you do it, anything
may happen. Some machines will crash, some will alter the literal, some
will overwrite [pseudo] random memory, some will write libellous emails
to your local head of state.
Some implementations have a stack, but `str[]` need not be on it.
Some implementations have read-only memory, but "" need not be in it.
But it is Not Allowed to update string literals.
--
Chris "try, or try not -- there is no do" Dollin
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