scanf函数的行为（） [英] Behaviour of scanf()

问题描述

  

可能重复：结果
  混乱（）与＆安培;运营商

我们为什么需要急症室;在scanf函数用于输入整数，为什么不为字符。
执行和放大器;在scanf函数是指merory位置同时获得输入。

例如： -

 的main（）
{
int类型的;
焦炭℃;scanf函数（％d个，＆安培; A）;
scanf函数（％CC）;
}
 

解决方案

对于每个转换说明， scanf（）的预计，相应的参数是一个指向正确的类型：％d个预计键入为int * ，％F 预计，键入双* ，％C 和％S 都期望类型的参数的char * 等

％C 和％S 的区别在于，前者让 scanf函数（ ）来读的单的字符并将其存储在由相应的参数指定的位置，而后者则讲述 scanf（）的读取多个字符，直到它看到一个0值字符并存储在缓冲区中开始参数所指定的位置的所有这些字符。

您需要使用＆安培; 运营商对你的论点，如果他们是不是已经指针类型。例如：

  INT X;
为int * PX = some_valid_memory_location_such_as_＆放大器; X;
焦炭℃;
字符* PC = some_valid_memory_location_such_as_和C;
...
scanf函数（％d个，＆安培; X）; // x不是一个指针类型，所以我们必须使用＆放大器;操作者
scanf函数（％D，像素）; //像素为指针类型，所以我们不需要和放大器;操作者
scanf函数（％C，和C）; //等等。
scanf函数（％C，PC）; //等等。
 

当事情变得扑朔迷离正在读的字符串的字符（使用％S 转换说明）：

 字符BUF [SIZE]
scanf函数（％S，BUF）; //前pression''BUF * *隐式转换为指针类型
 

为什么不我们需要的＆安培; 运营商在这种情况下？它用C如何处理数组前pressions做。当编译器看到数组类型的前pression（如 BUF 在 scanf（）的调用） ，它会隐式地从类型转换除权pression T N个元素的数组来指针至T ，和其值设置为阵列中的第一个元素的地址。这个值不是一个左值 - 它不能被分配​​到（所以你不能写东西像 BUF = foo的）。唯一的例外是当数组前pression要么是的sizeof 或＆放的操作; 运营商，或者如果数组前pression是一个字符串用来初始化另一个数组：

 的char * p =这是一个测试; //字符串隐式转换为char *，
                             //字符串*地址*写为p
所以char a [] =这是一个测试; //字符串不是隐式转换，
                             //字符串* *的内容复制到
 

总之，前pression BUF 是隐的从类型转换的char [SIZE] 到的char * ，所以我们并不需要使用＆安培; 运营商，并在事实上，前pression类型＆放大器; BUF 是字符指针的大小元素的数组，或（*）[SIZE] ，这是不是有什么 scanf（）的预计为 ％S 转换说明。

Possible Duplicate:
confusion in scanf() with & operator

Why we need a & in scanf for inputting integer and why not for characters. Do the & in scanf refers to merory location while getting input.

Eg:-

main()
{
int a;
char c;

scanf("%d",&a);
scanf("%c"c);
}

解决方案

For each conversion specifier, scanf() expects the corresponding argument to be a pointer to the proper type: %d expects an argument of type int *, %f expects an argument of type double *, %c and %s both expect an argument of type char *, etc.

The difference between %c and %s is that the former tells scanf() to read a single character and store it in the location specified by the corresponding argument, while the latter tells scanf() to read multiple characters until it sees a 0-valued character and store all those characters in the buffer starting at the location specified by the argument.

You need to use the & operator on your arguments if they are not already of pointer type. For example:

int x;
int *px = some_valid_memory_location_such_as_&x;
char c;
char *pc = some_valid_memory_location_such_as_&c;
...
scanf("%d", &x); // x is not a pointer type, so we must use the & operator
scanf("%d", px); // px is a pointer type, so we don't need the & operator
scanf("%c", &c); // etc.
scanf("%c", pc); // etc.

Where things get confusing is reading strings of characters (using the %s conversion specifier):

char buf[SIZE];
scanf("%s", buf);    // expression 'buf' *implicitly* converted to pointer type

Why don't we need the & operator in this case? It has to do with how C treats array expressions. When the compiler sees an expression of array type (such as buf in the scanf() call), it will implicitly convert the expression from type N-element array of T to pointer to T, and set its value to the address of the first element in the array. This value is not an lvalue -- it cannot be assigned to (so you can't write something like buf = foo). The only exceptions to this rule are when the array expression is an operand of either the sizeof or & operators, or if the array expression is a string literal being used to initialize another array:

char *p = "This is a test";  // string literal implicitly converted to char *,
                             // string *address* written to p
char a[] = "This is a test"; // string literal not implicitly converted,
                             // string *contents* copied to a

In short, the expression buf is implicitly converted from type char [SIZE] to char *, so we don't need to use the & operator, and in fact the type of the expression &buf would be pointer to SIZE-element array of char, or (*)[SIZE], which is not what scanf() expects for the %s conversion specifier.

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