0
Why there is no output of the following code?
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
8 ответов
+ 8
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!
+ 8
Hello, 😊
If you need help you can post the code you are struggling with!
• SEARCH for similar QUESTIONS or ANSWERS before posting
• Include relevant TAGS
• https://www.sololearn.com/post/75089/?ref=app
• https://code.sololearn.com/WvG0MJq2dQ6y/
+ 8
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!!
+ 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
+ 3
Your scanf() has four format specifiers (%d, %f etc.) for three variables
+ 3
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);
+ 2
Danijel, but what if we don't want to skip the input float field and still print the values of x and y?
+ 1
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.