C first fgets() is being skipped while the second runs

Question

The question is:

Why does the first fgets statement is being skipped? I read somewhere that it might be because of SCANF() that I used before. I am trying to figure it out but I can't. Can someone give me the solution (I should probably re-write the first bit of code to avoid scanf, but how?).

This in the code I am struggling with:

for(;;)
    {
            //Ask if the user wants to add another CD - Y/N
            fputs("\nWould you like to enter new CDs details?  y or n\n", stdout);
            scanf(" %c" ,&type);
            if (toupper(type) != 'Y')
                break;

            puts("");


            //getting in the album information        
            printf("\tLets enter the details of the CD %d:\n\n", count + 1);


            fputs("Title?\n", stdout);   

            //this fgets statement is being skipped
            fgets(title[count], sizeof title[count], stdin);
            title[count][strlen(title[count]) - 1] = '\0';

            fputs("Atrist? \n", stdout);
            fgets(artist[count], sizeof artist[count], stdin);
            artist[count][strlen(artist[count]) - 1] = '\0';
    }

Don't mix `scanf()` and `fgets()`. Use only `fgets()` for user input. — pmg, Jun 08 '15 at 10:48

Natasha Dutta · Accepted Answer · 2015-06-08T14:50:57.543

3

This is because the last ENTER keypress, which causes a newline is left in the input buffer. This is picked up by the first fgets().

You can add a while(getchar() != '\n'); before the first fegts() to avoid this.

[EDIT: Or, for better, as mentioned by Chux^{Thanks to him} in below comment,use something like

int ch; while((ch = getchar()) != '\n' && ch != EOF);

to handle the 'newline' as well as EOF.]

That said, it is never a good choice to mix scanf() and fgets(). Use fgets() always, it is possible and better.

edited Jun 08 '15 at 14:50

answered Jun 08 '15 at 10:49

Natasha Dutta

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Maybe `int ch; while((ch = getchar()) != '\n' && ch != EOF);` to prevent infinite loop of `while(getchar() != '\n');` on end-of-file. – chux - Reinstate Monica Jun 08 '15 at 14:37
@chux Done, updated. :-) – Natasha Dutta Jun 08 '15 at 14:51

score 1 · Answer 2 · answered Jun 08 '15 at 10:49

1

Yes, it's because your scanf() didn't read more than a single character, but the user pressed return. The return remains in the input buffer, so fgets() immediately sees that and returns.

Don't mix them, use only fgets().

answered Jun 08 '15 at 10:49

unwind

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score 0 · Answer 3 · answered Jun 08 '15 at 10:49

0

Simply change

scanf(" %c" ,&type);

to

scanf(" %c%*c" ,&type);

The reason that the first fgets gets skipped is that your scanf leaves a newline character in the stdin. fgets sees this character and consumes it, thus not waiting for furthur input.

The %*c tells scanf to read and discard a character.

answered Jun 08 '15 at 10:49

Spikatrix

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Thanks guys! Very helpful tips. I am going to look through the forum more carefully next time! – bartaspol Jun 13 '15 at 12:24

C first fgets() is being skipped while the second runs

3 Answers3