print 1 to 100 without using recursion and conditions

C (90) (79) (59) (47) (42) (40)

static int

x=1;a(){char b[8];printf("%d\n",x++);b[24]-=5*(1-x/101);}main(){a();return 0;}

The function a which prints the numbers does not call itself! I exploited a buffer overflow and changed the return address to make the program counter go over function a again as long as I need.

I don't know if this is considered to be a recursion, but I thought it would worth trying. This code works on my 64-bit machines with gcc 4.6, for other platforms the last statement of function a, could be a little different.

Exp1: I allocated a dummy buffer on stack b, and then addressed a passed-by-end location, which is the location of return address. I anticipated the distance between start of buffer and return address location from disassembly of function a.

Exp2: Expression 5*(1-x/101), is 5 for all x<=100 and 0 for x=101. By looking at disassembly of main (in my case), if you decrease the return address by 5, you will set the PC to calling point of a again. In the updated codes, the return value of printf is used for checking loop condition.

Update: After applying ugoren's suggestions and some other changes:

x;a(){int b[2];b[3*(printf("%d\n",++x)&2)]-=5;}main(){a();}

Update2: After Removing function a:

x;main(){int b[2];b[6^printf("%d ",++x)&4]-=7;}

Update3:

x;main(b){(&b)[1|printf("%d ",++x)&2]-=7;}

Update4: Thanks to mbz :)

x;main(b){(&b)[3|printf("%d ",++x)]-=7;}

85

C (gcc)

#define c printf("%d ",i++);
#define b c c c c c
#define a b b b b b
main(i){a a a a}

Assuming no command line arguments were passed.

C++ (159 136)

With templates.

#include<cstdio>
#define Z(A,B,C,D)template<A>struct P B{P(){C;printf("%d ",D);}};
Z(int N,,P<N-1>(),N)Z(,<1>,0,1)int main(){P<100>();}