const int x = 5; [英] const int x=5 ;
问题描述
可以告诉我为什么以下代码无法编译：
int main（无效）
{
const int a = 5;
int buf [a];
返回0;
}
语句中似乎有一些错误const int i = 5;
谢谢
Can anyboby please tell me that why the following code is''nt compiling:
int main(void)
{
const int a=5;
int buf[a];
return 0;
}
There seems to be some error in the statement const int i =5;
Thanks
推荐答案
ca ******** @ yahoo.com 写道：
可以告诉我为什么下面的代码没有编译：
int main（void）
{
const int a = 5;
int buf [a];
返回0;
}
语句const int i中似乎有一些错误5;
Can anyboby please tell me that why the following code is''nt compiling:
int main(void)
{
const int a=5;
int buf[a];
return 0;
}
There seems to be some error in the statement const int i =5;
首先，在C语言中''const int i = 5''不是声明，它是一个
声明。
其次，你的代码中没有''const int i = 5''。你可能意味着
''const int a = 5''。
第三，在C语言中，常量积分对象不符合
''整数常量''并且不能用于整数常量表达式
（ICE）。
用原始C语言（C89 / 90）数组声明中的数组大小必须是
是一个ICE。在你的代码中它不是。这就是为什么代码没有编译
（假设你使用的是C89 / 90编译器）。
在C99中这样的数组大小声明不需要是ICE，
意味着在C99中这段代码是合法的，''buf''将成为
可变长度数组（VLA） 。
-
祝你好运，
Andrey Tarasevich
Firstly, in C language ''const int i =5'' is not a statement, it is a
declaration.
Secondly, there''s no ''const int i =5'' in your code. You probably meant
''const int a=5''.
Thirdly, in C language constant integral objects do not qualify as
''integer constants'' and cannot be used in integral constant expressions
(ICE).
In the original C language (C89/90) array size in array declaration must
be an ICE. In your code it is not. That''s why the code is not compiling
(assuming that you are using a C89/90 compiler).
In C99 the array size in such declaration is not required to be an ICE,
meaning that in C99 this code would be legal and ''buf'' would become a
variable-length array (VLA).
--
Best regards,
Andrey Tarasevich
ca********@yahoo.com 写道：
ca********@yahoo.com wrote:
可以告诉我为什么下面的代码没有编译：
int main（void）{
const int a = 5;
int buf [a];
返回0;
}
在语句const int i = 5中似乎有一些错误;
cat main.c
#include< stdio.h>
int main（void）{
const
int n = 5;
int buf [n];
for（int j = 0; j< n; ++ j）
buf [j] = j;
for（int j = 0; j< n; ++ j）
fprintf（stdout，" ; buf [％d] =％d \ n"，j，buf [j]）;
返回0;
}
gcc -Wall -std = c99 -pedantic -o main main.c
./main
Can anyboby please tell me that why the following code is''nt compiling:
int main(void) {
const int a = 5;
int buf[a];
return 0;
}
There seems to be some error in the statement const int i = 5; cat main.c #include <stdio.h>
int main(void) {
const
int n = 5;
int buf[n];
for (int j = 0; j < n; ++j)
buf[j] = j;
for (int j = 0; j < n; ++j)
fprintf(stdout, "buf[%d] = %d\n", j, buf[j]);
return 0;
}
gcc -Wall -std=c99 -pedantic -o main main.c
./main
buf [0] = 0
buf [ 1] = 1
buf [2] = 2
buf [3] = 3
buf [4] = 4
它似乎对我来说很好。
buf[0] = 0
buf[1] = 1
buf[2] = 2
buf[3] = 3
buf[4] = 4
It seems to work just fine for me.
Andrey Tarasevich写道：
Andrey Tarasevich wrote:
ca ******** @ yahoo.com 写道：
ca********@yahoo.com wrote:
可以告诉我为什么下面的代码没有编译：
int main（void）{
const int a = 5;
int buf [a] ;
返回0;
语句const int a = 5;
Can anyboby please tell me that why the following code is''nt compiling:
int main(void) {
const int a = 5;
int buf[a];
return 0;
}
There seems to be some error in the statement const int a = 5;
在C语言[标准文档]，
''中似乎有一些错误const int a = 5''不是声明，它是声明。
In C language [standards documents],
''const int a = 5'' is not a statement, it is a declaration.
它*是*一个声明。声明*是*声明。
声明与语句区分
仅在ANSI / ISO C标准文档的上下文中。
It *is* a statement. A declaration *is* a statement.
Declarations are distinguished from statements
only in the context of the ANSI/ISO C standards documents.
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