scanf and whitespace characters

Question

I am doing a simple example related to scanf()

int a, b;
printf("Please enter int a:\n);
scanf("%d",&a);//line 1
printf("Please enter int b:\n);
scanf("%d",&b);//line 2

I run this code at a I put 45, and b 78, so the input buffer looks like this: 45\n78\n, line1 takes 45 ignores \n, and line2 ignores \n takes 78 and ignores \n

char ch, ch2;
printf("Please enter char ch:\n);
scanf("%c",&ch);//line3
printf("Please enter char ch2:\n);
scanf("%c",&ch2);//line4

I debugged this code and I thought of putting 'a' in ch, and 'b' in ch2, so the input buffer will look like this: a\nb\n, line3 takes 'a' ignores \n and line4 ignores \n takes 'b' and ignores \n

I thought this would happen, but when I debugged it line3 takes 'a', and line4 reads \n and stores it.

I don't understand I thought scanf() is supposed to ignore whitespace characters.

as you can see here http://www.cplusplus.com/reference/cstdio/scanf/

Note that you are missing corresponding double quotes in your code. — machine_1, Mar 07 '18 at 09:43
@machine_1 yeah that is only in the post not in the code, I will edit it right away — Embedded_Dude, Mar 07 '18 at 09:52
The `%c`, `%[]`, and `%n` directives do _not_ skip leading whitespace characters. — ad absurdum, Mar 07 '18 at 09:53

score 3 · Accepted Answer · edited Jun 20 '20 at 09:12

3

From man scanf:

%c

Matches a sequence of characters whose length is specified by the maximum field width (default 1); the next pointer must be a pointer to char, and there must be enough room for all the characters (no terminating null byte is added). The usual skip of leading white space is suppressed. To skip white space first, use an explicit space in the format.

edited Jun 20 '20 at 09:12

Community

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answered Mar 07 '18 at 09:35

Mathieu

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makes sense, the %c specifier won't suppress leading whitespace characters. Thanks for the comment – Embedded_Dude Mar 07 '18 at 09:53

scanf and whitespace characters

1 Answers1