在屏幕上打印19 [英] print 19 on screen
问题描述
#include< stdio.h>
int main()
{
char p =''19 '';
-----------------
返回0;
}
任何人都可以通过写一下19
将在屏幕上打印来告诉我什么是遗漏的陈述???
#include <stdio.h>
int main()
{
char p=''19'';
-----------------
return 0;
}
can anyone tell me what is the missing statement by writing which 19
will be printed on screen ???
推荐答案
asit写道:
asit wrote:
#include< stdio.h>
int main()
{
char p ='''19'';
#include <stdio.h>
int main()
{
char p=''19'';
您对此有何看法?
What do you expect this to do?
1月14日下午6:19,Mark Bluemel < mark_blue ... @ pobox.comwrote:
On Jan 14, 6:19 pm, Mark Bluemel <mark_blue...@pobox.comwrote:
asit写道:
asit wrote:
#include< stdio.h中>
#include <stdio.h>
int main()
{
char p ='''19'';
int main()
{
char p=''19'';
//写一个声明在屏幕上打印19
返回0;
}
//write a statement to print 19 on screen
return 0;
}
文章< 33 ************************** ********@i3g2000hsf.g ooglegroups.com>,
asit< li ***** @ gmail.comwrote:
In article <33**********************************@i3g2000hsf.g ooglegroups.com>,
asit <li*****@gmail.comwrote:
> #include< stdio.h>
int main()
char p ='''19'';
-----------------
返回0;
}
任何人都可以通过写下哪些19
将在屏幕上打印,告诉我什么是缺失的陈述?
>#include <stdio.h>
int main()
{
char p=''19'';
-----------------
return 0;
}
can anyone tell me what is the missing statement by writing which 19
will be printed on screen ???
你可以尝试
printf(" 19 \ nn);
也许你的意思是前一行是
char * p =" 19"
在这种情况下
printf("%s \ n",p);
可行。
- 理查德
-
:wq
You could try
printf("19\n");
Perhaps you meant the preceding line to be
char *p="19";
in which case
printf("%s\n", p);
would work.
-- Richard
--
:wq
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