一元 '*' 的类型参数无效（有“int”）

发布于 2024-12-10 09:08:30 字数 636 浏览 1 评论 0原文

我正在解决科学家和工程师的 C 的以下作业问题：

Given the following declarations and assignments, what do these expressions evaluate to?

int a1[10] = {9,8,7,6,5,4,3,2,1}
int *p1, *p2;
p1 = a1+3;
Line 14: p2 = *a1[2];

我正在尝试使用 gcc 编译此代码，但是当我这样做时，它会给出以下错误：

w03_3_prob15.c: In function 'main':
w03_3_prob15.c:14:7: error: invalid type argument of unary '*' (have 'int')

我正在使用以下命令进行编译：

gcc -o w03_3_prob15 w03_3_prob15.c -std=c99

我真的不知道该怎么做。您对如何修复此错误有任何想法吗？

原文

I'm working on the following homework problem from C for Scientists and Engineers:

Given the following declarations and assignments, what do these expressions evaluate to?

int a1[10] = {9,8,7,6,5,4,3,2,1}
int *p1, *p2;
p1 = a1+3;
Line 14: p2 = *a1[2];

I'm trying to compile this code with gcc, but when I do so, it gives me the following error:

w03_3_prob15.c: In function 'main':
w03_3_prob15.c:14:7: error: invalid type argument of unary '*' (have 'int')

I'm using the following command to compile:

gcc -o w03_3_prob15 w03_3_prob15.c -std=c99

I'm really not sure what to do. Do you have any ideas on how to fix this error?

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给我一枪 2024-12-17 09:08:30

该行无法编译，因为它在书中是不正确的。来自作者的勘误页面：

Page 438, line 17 from the bottom.
p2 = *a1[2]; 
should be  p2 = &a1[2];

The line doesn't compile because it's incorrect in the book. From the author's Errata page:

Page 438, line 17 from the bottom.
p2 = *a1[2]; 
should be  p2 = &a1[2];

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或十年 2024-12-17 09:08:30

p2 = *a1[2];

在 C 中，一元 * 仅为指针定义。 p2 是一个int*。 a1 是一个int[]。 a1[2] 是一个int。 [] 的优先级高于一元 *，因此您有 *(a1[2])，它不是合法的表达式。这就是编译器停止的原因。

我可以想到两种可能的解决方案。您想要哪一个，取决于您想要做什么。

*p2 = a1[2]; // Assigns the value of the second int in the array to the location
             // pointed to by p2.
p2 = &a1[2]; // Assigns the location of the second int in the array to p2.

p2 = *a1[2];

In C, unary * is only defined for pointers. p2 is an int*. a1 is an int[]. a1[2] is an int. [] has higher precedence than unary *, so you have *(a1[2]), which is not a legal expression. This is why the compiler is halting.

I can think of two possible solutions for this. Which one you want, depends on what you are trying to do.

*p2 = a1[2]; // Assigns the value of the second int in the array to the location
             // pointed to by p2.
p2 = &a1[2]; // Assigns the location of the second int in the array to p2.

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