I will save some poor old soul in here who uses malloc to allocate memory who will inevitably start to program on 64-bit machines. maybe you already do at work and you use the new keyword instead of malloc but after a day or so of messing with this problem i will save somoene the headache. i was porting a 32-bit app written in C(not written by me) to a 64-bit machine. everything should work fine right? WRONG! heres a snippit of code on a 64-bit machine: #include <stdio.h> #include <math.h> float *dat; int x = 100; int y = 200; {.... ..... //...do some stuff dat = (float *) malloc (sizeof(float) * x * y); printf("%f\n", dat[0]); ..... .... return 1; } so you would think that you would get an echo of whatever was in dat[0] right? i mean even if it's junk you should get something. no, actually you dont, you get a segmentation fault, ok that is half the answer.. the other answer is why? before i give the answer maybe someone can take a guess as to what is wrong. i am not a C programmer so ive always used the new keyword so ive never ran into this problem. but i will tell you, you learn A LOT of stuff just by porting programs to and from 32/64 bit machines. ok have at it :)

you still will get a seg fault, i changed it to %f though. i have already solved the problem, i was wondering if anyone else could figure it out. the 64-bit compiler will indeed give you a warning at this point..but thats all you will get...this one is a bit nasty for those that are C++ programmers. a C programmer in here might pick up on this ...:)

any number.. say 100 for x and 200 for y

float* will give a 64 bit pointer and I guess malloc returns a 64 bit pointer and printf by default thinks it is looking at a 32 bit value...

Quote:Original post by Trip99
float* will give a 64 bit pointer and I guess malloc returns a 64 bit pointer and printf by default thinks it is looking at a 32 bit value...

tripp you ALMOST got it...haha thats CLOSE..

Would this work:

printf("%fI64\n",dat[0]); ?

no scrub that - hmmmmm