#include <stdio.h>
int main()
{
printf("Hello, World!\n");
return(0);
}I very much agree with this post.
Forget party hats, Christmas tree hats all the way man.
When I tried compiling it, I got an error message saying that stdio.h doesn't exist! What ever could be wrong?
Install libc-dev.
This got me wondering, though. If libc-dev is for source code compilation, then what's the regular libc for?
Well, your 0 in "return (0)" should not be in parenthesis. Other than that, I don't know.
Yeah C never implicitly casts to boolean unless you do something like bool x = 1;
Hello, I suck at computers :)
And even then, bool is a C++ type, not C. C does not have a boolean type.
Drama, too long, didn't read, lol.
unsigned int fib(unsigned int n){
if (n < 2)
return n;
else
return fib(n - 1) + fib(n - 2);
}int main()
{
int fib(int x)
{
if x<2
return 1;
else
return fib(x-1)+fib(x-2);
}
int n;
printf("Integer goes here, yo: ");
scanf("%i", &n);
printf("\nThe Fibbonacci number of %i is %i", n, fib(n));
return 0;
}I was surfing the internet when I came upon this little thing:I borrowed it for use in my own code.unsigned int fib(unsigned int n){ if (n < 2) return n; else return fib(n - 1) + fib(n - 2); }
The program fails when trying to print the numbers. Why?int main() { int fib(int x) { if x<2 return 1; else return fib(x-1)+fib(x-2); } int n; printf("Integer goes here, yo: "); scanf("%i", &n); printf("\nThe Fibbonacci number of %i is %i", n, fib(n)); return 0; }
More troubles! I was surfing the internet when I came upon this little thing: <snip> I borrowed it for use in my own code. And this is how my code looks at the moment: <snip> The program fails when trying to print the numbers. Why? (Because %i signifies an integer, right? So I have to use %i two times, right?) And would it be possible to put the \n line break in the scanf, to make the code neater? And what's up with signed/unsigned integers? What does it mean?
Binary SignedValue UnsignedValueSo you see, if an integer is "signed", it sacrifices half of its range (i.e. one bit) to represent negative values; whereas if it is "unsigned", it uses the full range (i.e. all bits) for positive values. "int" is generally 32 bits, so its unsigned range is 0..(2^32-1) and its signed range is -(2^31)..(2^31-1)
0000 0 0
0001 1 1
0010 2 2
0011 3 3
0100 4 4
0101 5 5
0110 6 6
0111 7 7
1000 -8 8
1001 -7 9
1010 -6 10
1011 -5 11
1100 -4 12
1101 -3 13
1110 -2 14
1111 -1 15
this is how my code looks at the moment:The program fails when trying to print the numbers. Why? (Because %i signifies an integer, right? So I have to use %i two times, right?) And would it be possible to put the \n line break in the scanf, to make the code neater? And what's up with signed/unsigned integers? What does it mean?int main() { int fib(int x) { if x<2 return 1; else return fib(x-1)+fib(x-2); } int n; printf("Integer goes here, yo: "); scanf("%i", &n); printf("\nThe Fibbonacci number of %i is %i", n, fib(n)); return 0; }
firstly, it's %d, not %i. http://www.cplusplus.com/reference/clibrary/cstdio/printf.html provides a nice reference for all the format identifiers.
Because you are not printing a "\n" at the end, and your shell prompt is probably overwriting what it is printing.
Your code has a syntax error. The if clause should have parentheses in it. Other than that, it looks okay.
#include <stdio.h>
int main()
{
int fib(int x)
{
if(x<2)
return 1;
else
return fib(x-1)+fib(x-2);
}
int n;
printf("Integer goes here, yo: ");
scanf("%i", n);
printf("\nThe Fibbonacci number of %i is %i\n", n, fib(n)); //I really believe that this line is faulty in the "...\n", n, fib(n))" declaration.
return 0;
}You shouldn't, because that's the most inefficient possible way of calculating fibonacci numbers. It's a classical example of using complex recursion (ie. more complicated than simple tail recursion), but also a classical example of where a recursive solution can lead to extremely inefficient results.
Also, if that code block is accurate, I'm surprised it gets to the printf before choking. The fib function definition should go above main, not inside it.