How to loop and increase by 0.01 everytime?

I'm really confused on this code.

Here's what I want it to do: Start with a "v" value of 5, carry out the rest of the functions/calculations, increase the "v" value by 0.01, carry out the functions/calculations, then increase the "v" value by 0.01 again, carry out the functions...do this 500 times or until a "v" value of 10.00 is reached, whichever is easier to code.

Here is my code at the moment:

//start loop over v  
for(iv=5;iv<=500;iv++) {  
    v=0.01*iv;
    //Lots and lots of calculations with v here
}

Here is what I get: I tried setting iv<=10 so it does 10 loops only just so I could test it first before leaving it on all night. It did only 6 loops, starting at v=0.05 and ending at 0.1. So the problem is that a) it didn't 开发者_运维百科run for 10 loops, b) it didn't start at 5.00, it started at 0.05.

Any help would be appreciated.

EDIT: Holy crap, so many answers! I've tried 2 different answers so far, both work! I've been staring at this and changing code around for 3 hours, can't believe it was so easy.

---

You need to start at iv = 500. and if you want 10 loops, and iv++ is the update, then you stop before 510.

Reason: v = 0.01*iv, so v = 5 means iv = 5/0.01 = 500. As for the number of iterations, if your for loop is of the form for (x = N; x < M; x++) (constant N and M), then max(0, M-N) loops are executed, if x is not changed in the loop and no weird stuff (e.g. overflow, hidden casts of negative numbers to unsigned, etc.) occurs.

EDIT

Instead of using v = 0.01 * iv, v = iv / 100.0 is probably more accurate. Reason: 0.01 is not exactly representable in floating point, but 100.0 is.

---

Changing SiegeX's code so it uses integers ("more accurate"):

double dv;
int iv;
for(iv = 500; dv <= 1000; iv += 1)
{
    dv = (double)iv / 100.0;
}

---

double iv;
for(iv = 5.0; iv <= 10.0 ; iv += 0.01) {
/* stuff here */
}

---

int i;
double v;
v = 5;
for (i = 0; i < 500; i++)
{
    v += 0.01;
    // Do Calculations Here.

    if (v >= 10.00) break;
}

This gives you both. This will iterate at most 500 times, but will break out of that loop if the v value reaches (or exceeds) 10.00.

If you wanted only one or the other:

The 10.00 Version:

double v;
v = 5.0;
while ( v < 10.00 )
{
    v += 0.01;
    // Do Calculations Here. 
}

The 500 iterations version:

double v;
int i;
v = 5.0;
for( i = 0; i < 500; i++ )
{
    v += 0.01;
    // Do Calculations.
}

(Note that this isn't C99, which allows for a cleaner declaration syntax in the loops).

---

iv <= 10 doesn't do it for 10 loops, it does it until iv is greater than 10.

---

//start loop over v  
for(iv=0;iv<500;iv++) //loop from 0 to 499
{  
    v=v+0.01; //increase v by 0.01
    //Lots and lots of calculations with v here
}

this should do it