I'm a newbie and just starting to learn C
I tried to use scanf for the first time:
I enter => rate= .04Code:#include <stdio.h> #include <stdlib.h> int main() { double rate; printf("how high is the rate : "); scanf("%f", &rate); printf("rate is %f\n", rate); return 0; }
I got => rate is -9255596313 etc.... instead of 0.04
What is wrong? FYI I'm using Visual C++ 2008 as compiler and the name of the program ends with .c
The format specifier for scanf() to read a double is actually %lf, not %f (which is for float).
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Okay, I changed %f to %lf
I enter => rate= .04
I got => rate is (blank)
On a hunch: printf's format specifier should remain as %f. I did not tell you to change it, but maybe you did and Microsoft's compilers' poor support for C99 is in effect.
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THANK YOU!!
I did change that =)
Now it works well. May I ask why it should stay %f at printf while it's %lf at scanf?
printf received float var which is by default promoted to double, so float and double have the same format
scanf receives float* which is different from double*
so float* and double* need to have different format specifiers in scanf (the function needs to know how many bytes of memory are available for writing)
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
That is just the way it works: for double, %lf is for scanf and %f is for printf.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
Thank you for the help everyone.
All is clear now
