C语言 C中一元'*'的无效类型参数（有'int'）错误
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/42875580/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Invalid type argument of unary '*' (have 'int') Error in C
提问by Imdad
#include <stdio.h>
#include <stdlib.h>
/*
*
*/
int main() {
int a[] = {5, 15, 34, 54, 14, 2, 52, 72};
int p = &a[1];
int q = &a[5];
printf(*(p+3));
printf(*(q-3));
printf(*(q-p));
printf(*p<*q);
return (EXIT_SUCCESS);
}
Errors: "initialization makes integer from pointer without a cast [-Wint-conversion]" and "invalid type argument of unary '*' (have 'int')". First error is shown twice for the initialisation of variables above. Second error is shown for each print statement.
错误：“初始化从指针生成整数而不进行强制转换 [-Wint-conversion]”和“一元 '*' 的无效类型参数（有 'int'）”。对于上述变量的初始化，第一个错误显示了两次。每个打印语句都会显示第二个错误。
I'm not sure what is going wrong, anyone know how I can fix this?
我不确定出了什么问题，有人知道我该如何解决这个问题吗？
回答by VHS
You forgot to make pand qintpointers. Also, you forgot to use the format specifier in the printfstatements. Try the following:
你忘了做p和qint指点。此外，您忘记在printf语句中使用格式说明符。请尝试以下操作：
#include <stdio.h>
#include <stdlib.h>
/*
*
*/
int main() {
int a[] = {5, 15, 34, 54, 14, 2, 52, 72};
int *p = &a[1];
int *q = &a[5];
printf("%d\n", *(p+3));
printf("%d\n", *(q-3));
printf("%d\n", *q-*p);
printf("%d\n", *p<*q);
return (EXIT_SUCCESS);
}
回答by Mark Segal
&a[3](or &a[5]) is a pointer type, i.e. int *.
&a[3](或&a[5]) 是指针类型，即int *.
pis defined as int.
p被定义为int。
So you need to define pand qas int *, like this:
所以你需要定义pand qas int *，像这样：
int * p = &a[1];
int * q = &a[5];
回答by Edward Karak
The unary operator &yields the address of its operand. The type is of T *, not T. Therefore you cannot assign a int *to an intwithout a cast. The expression
一元运算符&产生其操作数的地址。类型是T *，不是T。因此，您不能在没有强制int *转换的int情况下将 a 分配给 an 。表达方式
&a[1]
yields the address of a[1].
产生 的地址a[1]。
I think you mean to define the variables as pointers to int, not just ints.
我认为您的意思是将变量定义为指向 int 的指针，而不仅仅是 ints。
int *p = &a[1];
int *q = &a[5];
