scanf and "\n" escape sequence

Question

I do not understand the functioning of scanf instruction

int i;
scanf("\n");
printf("hello!\n");
scanf("%d",&i);
printf(%d",i);

so the the second scanf don't let me type the second input, and it take only the first value why ?

you might want to take a look at this: http://stackoverflow.com/questions/15443483/using-n-in-scanf-in-c — Yohanes Khosiawan 许先汉, Apr 10 '14 at 23:57
actually I have seen this post before I ask my question, but realy I don't understood — user3466199, Apr 11 '14 at 00:03

score 1 · Accepted Answer · answered Apr 11 '14 at 00:13

1

when you (for example) input 2[space][space][space], the whitespaces will be consumed by \n, even your "enter"! (when you actually want to confirm your input).. and your 2 will be stored in variable i..

this \n will keep scanning until it finds a non-whitespace character, such as: your second int input.. but of course it's not stored in any variable (see your code)!

so at last, it returns and print out the value of variable i; which is 2..

answered Apr 11 '14 at 00:13

Yohanes Khosiawan 许先汉

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but why it is stored directely in i variable whitout ask me to type the second input when it execute the second scanf – user3466199 Apr 11 '14 at 00:26
because your `2` is not consumed by `\n` (it only consumes non-whitespaces), so it's still there **to be consumed** by **the one who wants it**, in this case, your **second scanner**.. – Yohanes Khosiawan 许先汉 Apr 11 '14 at 00:33

scanf and "\n" escape sequence

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