如何输入超过1个字符并仍然逐步执行程序？ [英] How do I enter more than 1 character and still execute program step wise ?

问题描述

当我输入下面的代码并尝试输入我的名字时，其余的代码执行自己而不允许我输入任何数据但是当我为名称输入单个字符时，我可以逐个输入数据。这是为什么 ？

When I enter the code below and try to enter my name then rest of the codes executive themselves without allowing me to enter any data but when I enter a single character for name then I can input the data one by one. Why is that ?

#include <stdio.h>

int main() {
	
	char name;
	int age;
        char gen;
	
	printf("What is your name ? \n");
	scanf(" %c", &name);
	
	printf("What is your age ? \n");
	scanf(" %i", &age);
	
	printf("What is your gender ? \n");
	scanf(" %c", &gen);
	
	printf("Thank you for the information! :)");
	return 0;
}

我的尝试：



我尝试过使用char name [10];我认为它可能是存储问题，我也将10改为1000但它似乎不起作用。每当我输入超过1个字符时，程序执行代码本身的其余部分。

What I have tried:

I have tried using "char name[10];" thinking it might be the storage problem, I have also changed "10" to "1000" but it doesn't seem to work. Whenever I input more than 1 character, the program executes the rest of the code itself.

推荐答案

％c是单个字符的格式说明符，这就是你的意思用过的。 ％s是文本字符串的格式说明符，它是您应该用于名称的格式说明符。

%c is the format specifier for a single character, which is what you used. %s is the format specifier for a text string and it is the one you should be using for the name.

Quote：

我尝试过使用char name [10];认为它可能是存储问题，

I have tried using "char name[10];" thinking it might be the storage problem,

还存储问题。在 C 编程语言中， char 数据类型表示单个字符。另一方面，字符串（例如名称）是（ 0 已终止） char acters。例如，请参阅 C Programming Strings [ ^ ]。

There is also a storage problem. In C programming language, char data type represents a single character. On the other hand, strings (e.g. names) are (0 terminated) array of characters. See, for instance, C Programming Strings[^].

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