Input/Output scanset in c


#include<stdio.h> int main() { char str[50]={'\0'}; scanf("%[A-Z]s",str); printf("%s",str); return 0; }

1) Input:







2) Input:








In output 1, i expected the output as "WORLD" but it didnt give any outout. From output 2, i understood that this is working only if the first few characters are in upper case.

Can you please explain how it actually works?


When you do


It takes input as long as you enter upper-case letters. And since you set all the array to '\0', printf() will stop printing when it meets one.

Therefore, the first input is blank, and the second is printing until the end of the upper-case string.


<h3>Interpretation of scansets</h3>

When it is given helloWORLD, the conversion specification %[A-Z] fails immediately because the h is not an upper-case letter. Therefore, scanf() returns 0, indicating that it did not successfully convert anything. If you tested the return value, you'd know that.

When it is given HELLoworlD, the scanset matches the HELL and stops at the first o. The format string also attempts to match a literal s, but there's no way for scanf() to report that it fails to match that after matching HELL.

<hr /><h3>Buffer overflow</h3>

Note that %[A-Z] is in general dangerous (as is %s) because there is no constraint on the number of characters read. If you have:

char str[50];

then you should use:

if (scanf("%49[A-Z]", str) != 1) ...some problem in the scan...

Also note that there is a 'difference by one' between the declared length of str and the number in the format string. This is awkward; there's no way to provide that number as an argument to scanf() separate from the format string (unlike printf()), so you may end up creating the format string on the fly:

int scan_upper(char *buffer, size_t buflen) { char format[16]; if (buflen < 2) return EOF; // Or other error indication snprintf(format, sizeof(format), "%%%zu[A-Z]", buflen-1); // Check this too!? return scanf(format, buffer); }


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