Why there is no output of the following code?

Question

scanf("%2d%d%f%5s", &x, &y, text);
printf ("%d %d %s", x, y, text) ;

// where x and y are integer variables and text is character type array
We've input as 1234 5.7 elephant

Answer 1

Krishna Kumar
Then, you don't need that input field!

⟹ Make sure order of format and variable data types match up.

Go through the lesson again, you'll better understand and use Daljeet Singh 's advice!

Answer 2

Hello, 😊
If you need help you can post the code you are struggling with!

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Answer 3

Krishna Kumar
You should correct the scanf() function for floating data type: %f

scanf("%2d %d %*f %5s", &x, &y, text);

• A format specifier can include several options along with a conversion character:

%[*][max_field]conversion character

⟹ The optional * will skip the input field!!

Answer 4

Krishna Kumar read about format specifiers here :
https://www.geeksforgeeks.org/format-specifiers-in-c/
in your code you are using %d instead use %s

// try inserting code by clicking @ "insert code"
https://code.sololearn.com/c3e7aVZ96l1P/?ref=app

Answer 5

Your scanf() has four format specifiers (%d, %f etc.) for three variables

Answer 6

Krishna Kumar add this (might come handy in future) to you code*
char text[5]="abcd";
printf("%8s\n",text);
printf ("%*s\n",0xf, text);

Answer 7

Danijel, but what if we don't want to skip the input float field and still print the values of x and y?

Answer 8

But here
scanf("%2d%d%f",&x,&y);
printf ("%d",x);
printf("\n%d",y);

//it also has greater format specifiers(3) as compared to lesser variables (2) but it gives output! Here x and y are also integer type.