sscanf(str1，“％s％d％f"，str2，#，&float1)产生意外结果 [英] sscanf(str1, "%s %d %f", str2,&num,&float1) producing unexpected results

问题描述

使用所包含的库，如何从存储在数组中的fget的输入中分离出包含名称，年龄和比率的字符串?

Using the included libraries, how do I separate a string containing a name, age, and rate, from the input from fgets stored in an array?

我需要修改字符串，年龄和格式化速率以显示3个小数位.之后，我需要将这些变量的新结果存储在单个char数组中.到目前为止，这就是我所拥有的.我不知道是什么原因导致sscanf()给我意外的结果或如何解决该问题

I need to modify the string, age, and format the rate to show 3 decimal places. After that, I need to store the new result of these variables in a single char array. This is what I have so far. I don't know what is causing sscanf() to give me the unexpected results or how to fix it

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <math.h>

#define SIZE 10
#define SIZE2 40

int main(){
char input[SIZE2]={};
char name[SIZE]={};
int age = 0;
float rate = 0;

printf("Enter name, age and rate (exit to quit): ");
fgets(input, SIZE2, stdin);

sscanf(input,"%s %d %f", name, &age, &rate);

printf("%s %d %.3f",name, &age, &rate);

return 0;
}

姓名显示正确，但是年龄是一个随机的大数字，比率显示为0.000

The name is displayed properly, but the age is some random large number and the rate is displayed as 0.000

推荐答案

删除printf(％s％d％.3f"，名称，& age，& rate)中的运算符(&)的地址；喜欢 printf(％s％d％.3f"，名称，年龄，费率)；

Remove the address of operator (&) in printf("%s %d %.3f",name, &age, &rate); like printf("%s %d %.3f",name, age, rate);

当您放置& printf()中的运算符打印变量的地址

When you put & operator in printf() It prints the address of the variable

但是为什么scanf()需要&运算符?

But why does scanf() require the & operator?

让我们看一个例子:

int main()
{
 int *ad,var;
 ad=&var;
 *ad=10;
 printf("%d",var);
 return 0;
}

它将打印10

让我们看看具有功能的另一个示例:

let see another Example with function:

void sum(int *sum,int a,int b)

int main()
{
 int a=10,b=10,c;
 sum(&c,a,b);
 printf("%d",c);
 return 0;
} 

void sum(int *sum,int a,int b)
{
 *sum=a+b;
}

它将打印20

因此，如果要在函数中修改变量，则可以通过引用传递变量

So,If you want to modify a variable in function You would pass the variable by reference

因此，scanf()需要&运算符以获取变量的地址

So,scanf() needs & of operator to get the address of variable

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