CODEWITHSUNDEEP
cioscanf
user445338
user445338

Reputation:

scanf not working. need to read double from console

I'm not sure what I'm doing wrong, but I'm not able to read a double from the console. Reading an it works fine for some reason. I'm using Xcode.

double n1;

// get input from the user
printf("Enter first number: ");
scanf("%f", &n1);

printf("%f", n1);

This will always print 0 no matter what I enter.

Upvotes: 6

Views: 8916

Answers (3)

Chokho
Chokho

Reputation: 103

Its all about how data is stored in memory.
Let me first tell that how long and float are stored in memory.

  • A double (long float, 64 bits) is stored in memory like or this (little endian notation).
  • Where as a float (32 bits) is stored like this (little endian notation).
    Also have a look at this "en.wikipedia.org/wiki/Floating_point#Internal_representation" (all floating data types)

So here you are asking to take input as %f (i.e. float, which is 4 bytes but double is 8 bytes), so compiler takes input from console and converts it to float type and stores it at memory location(which is actually 8 bytes) of variable (here n1).

Upvotes: 2

Richard J. Ross III
Richard J. Ross III

Reputation: 55543

%f is meant for a single precision floating-point value (float). The format specifier you need is %lf, meaning long precision floating-point value (double).

Upvotes: 6

Collin
Collin

Reputation: 12287

%f is looking for a float, not a double. If you want to use a double, use the format %lf.

As a somewhat interesting aside, clang warns about this without any extra flags, gcc 4.6 won't warn about it even with -Wall -Wextra -pedantic.

Upvotes: 6

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