字符串作为数组索引 [英] String as an array index

问题描述

在3["XoePhoenix"]中，数组索引是字符数组类型.我们可以用C语言做到吗?数组索引必须是整数不是真的吗?

In 3["XoePhoenix"], array index is of type array of characters. Can we do this in C? Isn't it true that an array index must be an integer?

3["XeoPhoenix"]是什么意思?

推荐答案

3["XoePhoenix"]与"XoePhoenix"[3]相同，因此它将求值为char 'P'.

3["XoePhoenix"] is the same as "XoePhoenix"[3], so it will evaluate to the char 'P'.

C中的数组语法与编写*( x + y )的方式完全不同，其中x和y是方括号之前和之内的子表达式.由于加法的可交换性，可以在不更改表达式含义的情况下交换这些子表达式.

The array syntax in C is not more than a different way of writing *( x + y ), where x and y are the sub expressions before and inside the brackets. Due to the commutativity of the addition these sub expressions can be exchanged without changing the meaning of the expression.

因此，3["XeoPhoenix"]被编译为*( 3 + "XeoPhoenix" )，其中字符串衰减为指针，而3添加到该指针，这又导致指向字符串中第4个字符的指针. *取消引用此指针，因此该表达式的值为'P'.

So 3["XeoPhoenix"] is compiled as *( 3 + "XeoPhoenix" ) where the string decays to a pointer and 3 is added to this pointer which in turn results in a pointer to the 4th char in the string. The * dereferences this pointer and so this expression evaluates to 'P'.

"XeoPhoenix"[ 3 ]将被编译为*( "XeoPhoenix" + 3 )，您会看到将导致相同的结果.

"XeoPhoenix"[ 3 ] would be compiled as *( "XeoPhoenix" + 3 ) and you can see that would lead to the same result.

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