C语言 如何打印有限数量的字符？
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How do you print a limited number of characters?
提问by Mike Pateras
Sorry to put a post up about something so simple, but I don't see what I'm doing wrong here.
很抱歉就这么简单的事情发表一篇文章，但我不明白我在这里做错了什么。
char data[1024];
DWORD numRead;
ReadFile(handle, data, 1024, &numRead, NULL);
if (numRead > 0)
printf(data, "%.5s");
My intention with the above is to read data from a file, and then only print out 5 characters. However, it prints out all 1024 characters, which is contrary to what I'm reading here. The goal, of course, is to do something like:
我上面的意图是从文件中读取数据，然后只打印出 5 个字符。但是，它打印出所有 1024 个字符，这与我在这里阅读的内容相反。当然，目标是执行以下操作：
printf(data, "%.*s", numRead);
What am I doing wrong here?
我在这里做错了什么？
回答by R Samuel Klatchko
You have your parameters in the wrong order. The should be written:
您的参数顺序错误。应该写成：
printf("%.5s", data);
printf("%.*s", numRead, data);
The first parameter to printfis the format specifier followed by all the arguments (which depend on your specifier).
第一个参数printf是格式说明符，后跟所有参数（取决于您的说明符）。
回答by AraK
I think you are switching the order of arguments to printf:
我认为您正在将参数顺序切换为printf：
printf("%.5s", data); // formatting string is the first parameter
回答by Brian Roach
You're not calling printf() correctly.
您没有正确调用 printf()。
int printf ( const char * format, ... );
Which means...
意思是...
printf("%.5s", data);
回答by sjchoi
You are using wrong syntax for the printfstatement, and the .number is only for numerical variables.
您对printf语句使用了错误的语法，并且 .number 仅用于数字变量。
So it should be
所以应该是
int i;
for(i=0;i<5;i++)
printf("%c", data[i]);
