什么会打印 - ％ - 30.4f [英] what will this print-%-30.4f

问题描述

i = 1234.5678;

printf（％ - 30.4，i）;



*这将打印什么 - ％ - 30.4f，在这个声明中是什么意思-30和如果它的+30和如果它的3等等*我知道它只是％.2f的含义它将只打印小数点后的2位数。

i=1234.5678;
printf("%-30.4",i);

*what will this print-%-30.4f ,in this statement whats the meaning of -30 and what if its +30 and what if its 3 etc. *I know the meaning if its just %.2f it will print only 2 digits after the decimal point.

推荐答案

这样的问题很容易找到答案。请先尝试为自己做一些基础研究。请参阅 https://msdn.microsoft.com/en-us/library/wc7014hz.aspx [ ^ ]。

Question like this are easy to find answers for. Please try and do at least some basic research for yourself first. See https://msdn.microsoft.com/en-us/library/wc7014hz.aspx[^].

Quote：

i = 1234.5678;

i=1234.5678;

您是否使用 i 作为名称一个浮点变量以欺骗敌人？ :-)

Do you use i as the name of a floating point variable in order to deceive the enemy? :-)

Quote：

printf （％-30.4，i）;

printf("%-30.4",i);

这是错误的：它缺少说明符（参见printfat C ++ reference [ ^ ]）。

你可能意味着

This is wrong: it lacks the specifier (see "printf" at C++ reference[^]).
You probably meant

printf("%-30.4f",i);

这意味着（就像你一样）可能很容易找到你自己，理查德已经告诉你了）：

• 在给定字段中左对齐（' - '）
• 字段总共使用30个字符
• 仅打印4位小数


That means (as you could have easy found yourself, Richard told you it already):


• Left align in the given field ('-')
• Use a total of 30 characters for the field
• print only 4 decimal digits




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