为什么我们不在printf中使用&符号（&）以及为什么我们在scanf中使用它？ [英] Why we dont use ampersand(&) in printf and why we use it in scanf ?

问题描述

在扫描f中未使用&符号（&）时，表明程序已停止工作。为什么我们只在c语言的scanf中使用它...请详细解释我&用语言简单说明



我尝试了什么：



当我是尝试运行程序没有&符号（&）然后显示，程序停止工作

解决方案

&是一个运算符，它解析为变量的地址，所以scan函数获取写入结果的地址（或指针）。



％是打印函数中的格式前缀。它就像一个标志，表示现在有特殊字符即将来临。



请阅读som基础编程教程。 ; - ）

在 C 编程语言中，函数参数按值按值。因此，被调用函数对其参数的任何修改都对调用者隐藏，例如尝试

  #include   <   stdio.h  >  
  
  void  square_it（ int  n）
 {
n = n * n; 
 printf（  square_it内，n =％d \ n，n）; 
} 
 
  int  main（）
 {
  int  i =  5 ;函数调用前
 printf（ ，i =％d \ n，i） ; 
 square_it（i）; 
 printf（ 函数调用后，i =％d \ n，i） ; 
  return   0 ; 
} 

所以， square_it 无法制作对调用者可见的更改。

指针是克服这种困难的有效方法：指针仍然按值传递（因此称为fucntion不能修改指针值），无论如何它们提供读取/ write 访问指向的值。尝试：

  #include   <   stdio.h  >  
  
  void  square_it2（ int  * p）
 {
 * p = * p * * p; 
 printf（  square_it2，* p =％d \ n，* p ）; 
} 
 
  int  main（）
 {
  int  i =  5 ;函数调用前
 printf（ ，i =％d \ n，i） ; 
 square_it2（& i）;  //  将poitner传递给i，即我的地址 
 printf（ ，i =％d \ n，i）; 
  return   0 ; 
} 

现在，虽然 printf 不需要更改其参数（它是一个输出函数） scanf 确实（它是一个输入函数）。因此它需要变量指针（地址）作为参数。

首先阅读C文档。

&的用法详细说明

&是关于你对变量的处理方法。

printf从变量中获取值

scanf将值写入变量并需要变量adres这样做

when not using ampersand(&) in scan f, then showing that the program has stopped working. why we use it only in scanf in c language.. please explain me in detail & in easy language

What I have tried:

when i am trying to run program without ampersand(&) then showing, program stoppted working

解决方案

The & is a operator, which resolves to "address of the variable", so the scan function gets the address (or pointer) to write the result.

The % is a formatting prefix in the print function. It is like a flag signaling that now special characters are coming.

Please read som fundamental programming tutorials. ;-)

In C programming language, function arguments are passe by value. Hence any modification the called function makes to its arguments is hidden to the caller, try for instance

 #include <stdio.h>

void square_it( int n )
{
  n = n * n;
  printf("inside square_it, n = %d\n", n);
}

int main()
{
  int i = 5;
  printf("before function call, i = %d\n", i);
  square_it( i );
  printf("after function call, i = %d\n", i);
  return 0;
} 

So, there's no way for square_it to make changes visible to the caller.
Pointers are an effective way to overcome such a difficulty: pointers are still passed by value (hence called fucntion cannot modify the pointer value), anyway they provide read/write access to the pointed value. Try:

 #include <stdio.h>

void square_it2( int * p )
{
  *p = *p * *p;
  printf("inside square_it2, *p = %d\n", *p);
}

int main()
{
  int i = 5;
  printf("before function call, i = %d\n", i);
  square_it2( &i ); // passing a poitner to i, that is i address
  printf("after function call, i = %d\n", i);
  return 0;
} 

Now, while printf doesn't need to change its arguments (it is an output function) scanf does (it is an input function). Hence it needs variable pointers (addresses) as arguments.

Start by reading C documentation.
The usage of & is explain in detail
& is about what you do with a variable.
printf get a value from a variable
scanf write a value to a variable and need the variable adres to do so.

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