scanf中的固定长度数组。格式警告

Question

So I have typedefed the following :

typedef char array[25];

In my code somewhere I call scanf like this :

scanf("%s",array);

But I get the warning that %s was expecting a char* while I pass a char*[25].Is there any way to get rid off this warning ?

Answer 1

array is type and is not a object. you can not pass it as argument in the scanf

array a;
scanf("%s", a);

Answer 2

typedef defines a data type so that you can use it to define different variables with same data types. like:

typedef char employee[25];
typedef char customer[25];
employee a, b;
customer y, z;
printf("Enter first employee's name: ");
scanf("%s",a);
printf("Enter customer's name: ");
scanf("%s",y);

though the datatypes are same, but defining them is elaborating things. all you need to do here is: take array a; and then scanf("%s",a);

Hope you are clear to the core! :)

Answer 3

I am assuming your code is abbreviated...

The actual reason for the warning is that the compiler knows two things: you have a char array that is limited length, and that scanf will scan as many characters as are given in the original string; potentially overflowing the array provided for them.

This is a real issue: the compiler is quite correct.

The solution is to specify the scanf format better. In this case I would suggest using "%24s" to tell scanf the max number of characters to read for that format element.

Remember to allow for the null terminator - better, explicitly zap array[len-1] with 0.

There are possibly better ways to read the string than scanf. Look into some of the token based methods, for example.

HTH, Ruth