scanf 和转换说明符的问题 [英] problems with scanf and conversion specifiers

问题描述

在这里你可以看到我的源代码:

Here you can see my source code:

#include <stdio.h>

int main()
{
    char yourname;
    char yoursex;
    int yourage = 0;

    printf("Hey, what's your name?\n");
    printf("My name is: ");
    scanf("%s", &yourname);
    printf("Oh, hello %s! \n\n", &yourname);

    printf("Are you a boy or a girl?: ");
    scanf("%s", &yoursex);
    printf("Nice to know you are a %s! \n\n", &yoursex);

    printf("How old are you %s? I am ", &yourname);
    scanf("%d", &yourage);
    printf("I see you are %d, you have many years then!", &yourage);

    return 0;
}

我正在尝试一些我不知道的东西，但奇怪的是它对我不起作用.有什么问题?另外，为什么它需要是 %s 而不是 %c?如果我改用 %c 则不起作用！

I was trying things that I didn't knew, and strangely it is not working for me. What's the problem? Also, why it needs to be %s and not %c? If I use %c instead it does not work!

它说:你 %s 几岁了?它没有写我的名字，而是写着oy"

Where it says: How old are you %s? instead of putting my name, it says ''oy''

最后一行没有显示我的年龄，而是显示了一个很大的数字.

and instead of showing my age in the last line, it shows a big number.

推荐答案

这些是 C 编程的基础知识，我强烈建议您买一本像样的书 - Dennis Ritchie 的 The C Programming Language 将是一个好的开始.

These are the very basics of C Programming, and I strongly advise you to get a decent book - The C Programming Language by Dennis Ritchie would be a good start.

您的代码中有很多错误.

There are numerous errors in your code.

1. 一个 char 只能包含一个字符，比如 'A' 或 'a' 或类似的东西.当您扫描姓名时，它将是一组字符，例如E"、d"、d"、y".要存储多个字符，您需要使用字符数组.另外，用于扫描/打印字符的格式说明符是 %c，%s 用于需要扫描一组字符时，也称为字符串到数组中.

1. A char can contain only one character, like 'A', or 'a' or something like that. When you're scanning a name, it is going to be a group of characters, like 'E', 'd', 'd', 'y'. To store multiple characters, you need to use a character array. Also, the format specifier used to scan/print characters is %c, %s is for when you need to scan a group of characters, also called a string into an array.

当您使用 printf 时，您没有提供指向要打印的变量的指针(&x 是指向变量 x).指针是一个 32/64 位整数，这可能是您在尝试打印时看到随机整数的原因.printf("%c\n", charVar) 就足够了.

When you use printf, you do not supply a pointer to the variable you are trying to print (&x is a pointer to variable x). The pointer is a 32/64-bit integer, which is likely why you see a random integer when trying to print. printf("%c\n", charVar) is sufficient.

scanf 在使用 %s 作为格式说明符时不需要 &，假设您已经传递了一个字符数组作为论据.原因是，scanf 需要知道在哪里存储您从输入中读取的数据——这是由指向内存位置的指针给出的.当你需要扫描一个整数时，你需要传递一个 &x - 这意味着，指向 x 的内存位置的指针.但是当你传递一个字符数组的时候，它已经是内存地址的形式了，不需要在前面加&.

scanf does not need an & while using %s as the format specifier, assuming you have passed a character array as the argument. The reason is, scanf needs to know where to store the data you are reading from the input - and that is given by a pointer to the memory location. When you need to scan an integer, you need to pass an &x - which means, pointer to memory location of x. But when you pass a character array, it is already in the form of a memory address, and doesn't need to be preceded by an ampersand.

我再次建议您在网上查找一些不错的教程，或者买一本书(我上面提到的那本书是经典之作).键入材料中给出的示例.实验.玩得开心.:)

I once again recommend you look up some decent tutorials online, or get a book (the one I mentioned above is a classic). Type the examples as given in the material. Experiment. Have fun. :)

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