谁能帮我怎么编译这个程序？ [英] can someone help me how to compile this program??
问题描述
以下是我正在尝试编译的程序
// isres.c
#include< stdio.h>
int main（）
{
char arr [10] =" Qualifiers"
char * restrict p = arr;
int i = 0;
for（; i< 10; ++ i）
printf（"％c"，arr [i]）;
返回0;
}
i我在窗口xp SP2中使用cygwin，我的gcc版本是3.4 .4
如何编译上面的程序？
i尝试使用这个
gcc -c -std = c99 restrict.c
但我收到这样的错误
restrict.c：在函数''main'中'：
restrict.c：6：错误：在char之前解析错误
有人可以帮助我如何编译这个程序吗？
The following is the program i am trying to compile
//restrict.c
#include <stdio.h>
int main()
{
char arr[10] = "Qualifiers"
char * restrict p = arr;
int i = 0;
for(; i < 10; ++i)
printf("%c", arr[i]);
return 0;
}
i am using cygwin in window xp SP2, my gcc version is 3.4.4
how do i compile the above program?
i tried using this
gcc -c -std=c99 restrict.c
but i get an error like this
restrict.c: In function ''main'':
restrict.c:6: error: parse error before "char"
can someone help me how to compile this program??
推荐答案
Anarki写道：
Anarki wrote:
以下是我试图编译的程序le
//resres.c
#include< stdio.h>
int main（）
{
char arr [10] =" Qualifiers"
The following is the program i am trying to compile
//restrict.c
#include <stdio.h>
int main()
{
char arr[10] = "Qualifiers"
缺少分号。
-
Ian Collins。
Missing semicolon.
--
Ian Collins.
char arr [10] =" Qualifiers"
你缺少;
试试：
char arr [10] =" Qualifiers";
char arr[10] = "Qualifiers"
you are missing ;
try:
char arr[10] = "Qualifiers";
10月1日，1：31 * pm，MisterE< mist ... @ no.email .okwrote：
On Oct 1, 1:31*pm, MisterE <mist...@no.email.okwrote:
* * char arr [10] =" Qualifiers"
* *char arr[10] = "Qualifiers"
你缺少;
试试：
char arr [10] =" Qualifiers" ;;
you are missing ;
try:
char arr[10] = "Qualifiers";
omg！我没看到***分号srry线程关闭了！
omg! i didnt see tht *** semicolon srry thread closed!
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