scanf 和 double 的问题 [英] Problems with scanf and doubles

问题描述

我无法理解为什么会发生这种情况:使用以下代码；

im having trouble understanding why this occurs: with the following code;

    #include <stdio.h>

    int main() 
    {
        double x=123;

        printf("x is %f. Enter a new value for x.\n", x);
        scanf("%f", &x);
        printf("x is %f\n", x);

        return 0;
    }

当您输入 45678 作为 x 的新值时，将打印x is 123.000017".我知道当你在扫描时使用 %lf 时这是固定的，但为什么在使用 %f 时会发生这种情况?

when you enter 45678 as the new value for x, 'x is 123.000017' is printed. I know that this is fixed when you use %lf when scanning, but why does this happen when using %f?

推荐答案

我知道当你在扫描时使用 %lf 时这是固定的，但为什么在使用 %f 时会发生这种情况?

I know that this is fixed when you use %lf when scanning, but why does this happen when using %f?

printf 和 scanf 函数参数之间的区别在于您将值传递给 printf，但传递给 scanf您传递指针(即值的地址).根据 C 规则，当一个函数接受可变数量的参数时，所有参数都会经历默认提升.

The difference between printf and scanf function arguments is that you pass values to printf, but to scanf you pass pointers (i.e. addresses of values). According to C rules, when a function takes a variable number of arguments, all parameters undergo default promotions.

其中一项促销是 float 转换为 double，与 short 转换为 的方式相同整数.这就是为什么您可以将 %f 或 %lf 与 printf 一起使用:这两种类型的值都被编译器转换为 double，因此访问它们是安全的.

One of the promotions is that floats get converted to double, in the same way that shorts get converted to int. That is why you can use %f or %lf with printf: both types of values are converted by the compiler to double, so accessing them is safe.

但是，指针没有这样的规则(并且有一个很好的理由:尝试将 double 写入 float 的空间将是未定义的行为).这就是为什么在将参数传递给 scanf 系列函数时必须区分 %f 和 %lf 的原因.

However, there is no such rule for pointers (and for a good reason: trying to write a double into a space of float would be undefined behavior). That is why you must differentiate between %f and %lf when passing parameters to the scanf family of functions.

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