指针增量和间接引用（左值所需错误） [英] pointer increment and dereference (lvalue required error)

问题描述

我想了解指针递增和提领走在一起，我这样做是为了尝试一下：

 的#include＆LT;＆stdio.h中GT;
INT主（INT ARGC，CHAR *的argv []）
{
    字符*字[] = {字词1，字词2};
    的printf（％P \\ N字）;
    的printf（％S \\ n，*字++）;
    的printf（％P \\ N字）;
    返回0;
}
 

我预计在code做的其中之一：

1. 先解引用再增加指针（印刷字1）


2. 先解引用然后增加值（印刷ord1）


3. 取消引用指针+ 1（印刷字词2）

但是编译器甚至不会编译这一点，并给出了此错误：<？code>左值要求作为增量的操作数我在这里做得不对<​​/ P>

解决方案

您不能增加一个数组，但你可以增加一个指针。如果转换声明一个指针数组，你会得到它的工作：

 的#include＆LT;＆stdio.h中GT;
INT主（INT ARGC，CHAR *的argv []）
{
    为const char * WW [] = {字词1，字词2};
    为const char **字= WW;
    的printf（％P \\ N字）;
    的printf（％S \\ n，*字++）;
    的printf（％P \\ N字）;
    返回0;
}
 

I am trying to understand how pointer incrementing and dereferencing go together, and I did this to try it out:

#include <stdio.h>
int main(int argc, char *argv[])
{
    char *words[] = {"word1","word2"};
    printf("%p\n",words);
    printf("%s\n",*words++);
    printf("%p\n",words);
    return 0;
}

I expected this code to do one of these:

1. First dereference then increase the pointer (printing word1)
2. First dereference then increase the value (printing ord1)
3. Dereference pointer + 1 (printing word2)

But compiler won't even compile this, and gives this error: lvalue required as increment operand am I doing something wrong here?

解决方案

You cannot increment an array, but you can increment a pointer. If you convert the array you declare to a pointer, you will get it to work:

#include <stdio.h>
int main(int argc, char *argv[])
{
    const char *ww[] = {"word1","word2"};
    const char **words = ww;
    printf("%p\n",words);
    printf("%s\n",*words++);
    printf("%p\n",words);
    return 0;
}

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