scanf（“％s”，s）;会发生什么？ 遇到两个或两个以上的单词，并且只提供了一个变量？

Question

When a scanf("%s",s); (one of many ways to get a string which is not perfect) encounters a space in the input, it will try to put it in another variable, right? But what hapens if there is only one variable provided as in this case?

Also what other ways are used to input a string? which is the esiest or best one to use and which one does not give problems like the gets(s); function?

Here is my s_insert function now:

// pointer to pointer needed when you allocate memory in the function
void s_insert(char **string_one){   //inserts string (loss of original data)
    *string_one=(char*)malloc(200);
    fgets (*string_one,200,stdin);
} 

Answer 1

scanf for %s data specification reads characters before first space symbol (' ', '\\n' or '\\t'). If you want to read string with spaces (more than two words) use fgets function, that is more safe than 'gets' because you can set the maximum number of character that can be allocated in your memory and avoid segmentation fault.

Answer 2

No, it will only try to "put in a variable" when it encounters a % with a suitable conversion specifier in the first argument. That argument is what controls its behavior, not the input.

Having a single %s and multiple words in the input will simply leave the remaining words still in the input buffer, since scanf() will stop when it's done with the single %s , it has nothing more to do then.

It reads its conversion specification string and tries to read input to match that, not the other way around.