Hi,
My build went fine and printed out in the console just fine. But the final line was “Program ended with exit code: -1”
Here was my code:
#include <stdio.h>
int main(int argc, const char * argv[])
{
// Declare a variable
float shoe;
// Assign number
shoe = 9.5;
float shoe2;
shoe2 = 7.5;
// Declare a variable of type double
double pair;
pair = shoe + shoe2;
printf("Size of J & T's feet = %f size. \n", pair);
return 0;
}
Hey Billygoat,
I copied the code exactly as you have it on the forum and everything looks correct.
Size of J & T’s feet = 17.000000 size.
Program ended with exit code: 0
Can you try rerunning the command and circle back on the forum?
Thanks
Hi…
When I try my code, I get an error which I am having trouble looking up. I get: "passing ‘double’ to parameter of incompatible type ‘const. char’"
Can someone help explain what this error means? Also can someone help me figure out how I invoked it? Thank you.
#include <stdio.h>
int main(int argc, const char * argv[]) {
float first = 3.14;
float second = 42.03;
double = first + second;
printf(third);
}
The problem I believe is that you don’t have a variable name “third” after the word “double”.
[quote]#include <stdio.h>
int main(int argc, const char * argv[]) {
float first = 3.14;
float second = 42.03;
double = first + second;
printf(third);
}
[/quote]
Look at this code:
[code]#include <stdio.h>
int main (int argc, const char * argv[]) {
float first = 3.14;
float second = 42.03;
double third = first + second;
printf ("%f", third);
}
[/code]
Can you see where you went wrong?
yea. It doesn’t look like the chapter explains this. Only the example code was given: printf(" Cook it for %f minutes.\ n", cookingTime); (Location 623) When I read this I thought “minutes” was the variable (my mistake) so I thought that the variable was expected to be within the quotations.
Anyway. Thank you.
The result concatenated with some random stuff:
45.169998Program ended with exit code: 0
Is there a way to exclude that and only get the result?
My code was this:
#include <stdio.h>
int main(int argc, const char * argv[]) {
float first = 3.14;
float second = 42.03;
double third = first + second;
printf("%f",third);
return 0;
}
If you put a newline character at the end of the format specifier, that will have the result on a line by itself.