scanf函数不打印或阅读什么 [英] scanf not printing or reading anything

问题描述

我有以下线在我的code：

I have the following line in my code:

char y[] = "a";
scanf("Letter: %s", y);
printf("%s\n", y);

第二行根本不影响第三线的输出。我已经包括＆LT;＆stdio.h中GT; ，我想不出有什么不好...

The second line does not impact the output of the third line at all. I've included <stdio.h>, I can't think of what is wrong...

推荐答案

一个人做的最大的错误，是包括在引号之间的 scanf函数函数的字符串比格式规范（如％S 或％d个）的$ C $其他C应该 scanf函数（％S，Y）。如果你采取任何其他字符，那么你将不得不从头开始你的头找，找出问题。

One of the biggest mistakes one does, is to include any string in the scanf function between the quotes other than format specification(like %s or %d).The code should be scanf("%s",y).If you take any other character then you will have to scratch your head looking to find out problems.

（即使你有任何字符，那么你必须输入该字符，EG-如果你写 scanf函数（信：％S，Y）; 然后上输入你必须写像 C：\\＆GT;信：的的信，你将进入的），这显然不是一个明智的想法。也在 scanf函数的功能是不是有打印出来的东西，它从terminal.To只是读取输入打印出来，你应该使用的printf（信 ）; 就是这样

(Even if you include any character then you have to input that character,e.g- if you write scanf("letter: %s",y); then on input you have to write like C:\>letter: "letter which you will enter") which is obviously not a wise idea .Also the scanf function is not there to print things out, its just to read input from the terminal.To print out you should use printf("letter"); that's it.

只是假设你有使用一个 scanf（）的则可以使用像 scanf函数（％D采取输入从两个int变量＆放大器;一，和b）; 你可以看到我已经把任何除引号格式规范

Just suppose you have to take inputs from two int variables using one scanf() then you will use like scanf("%d%d",&a,&b); as you can see i have put nothing except format specification in the quotes.

这篇关于scanf函数不打印或阅读什么的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！