使用scanf（）输入字符串

Question

I started learning about inputting character strings in C. In the following source code I get a character array of length 5.

#include<stdio.h>
int main(void)
{
    char s1[5];
    printf("enter text:\n");
    scanf("%s",s1);
    printf("\n%s\n",s1);
    return 0;

}

when the input is:

1. 1234567891234567 , and I've checked it's working fine up to 16 elements(which I don't understand because it is more than 5 elements).
2. 12345678912345678 , it's giving me an error segmentation fault: 11 (I gave 17 elements in this case)
3. 123456789123456789 , the error is Illegal instruction: 4 (I gave 18 elements in this case)

I don't understand why there are different errors. Is this the behavior of scanf() or character arrays in C?. The book that I am reading didn't have a clear explanation about these things. FYI I don't know anything about pointers. Any further explanation about this would be really helpful.

Answer 1

Is this the behavior of scanf() or character arrays in C?

TL;DR - No, you're facing the side-effects of undefined behavior .

To elaborate, in your case, against a code like

 scanf("%s",s1);

where you have defined

 char s1[5];

inputting anything more than 4 char will cause your program to venture into invalid memory area (past the allocated memory) which in turn invokes undefined behavior .

Once you hit UB, the behavior of the program cannot be predicted or justified in any way. It can do absolutely anything possible (or even impossible).

There is nothing inherent in the scanf() which stops you from reading overly long input and overrun the buffer, you should keep control on the input string scanning by using the field width, like

 scanf("%4s",s1);  //1 saved for terminating null

Answer 2

The scanf function when reading strings read up to the next white-space (eg newline, space, tab etc.), or the "end of file". It has no idea about the size of the buffer you provide it.

If the string you read is longer than the buffer provided, then it will write out of bounds, and you will have undefined behavior .

The simplest way to stop this is to provide a field length to the scanf format, as in

char s1[5];
scanf("%4s",s1);

Note that I use 4 as field length, as there needs to be space for the string terminator as well.

You can also use the "secure" scanf_s for which you need to provide the buffer size as an argument:

char s1[5];
scanf_s("%s", s1, sizeof(s1));