unkown number of nested for loops

[Klaas Vantournhout]

Hi, Somehow I can't figure this one out.

I have an integer variable N and this N reflects the amount of nested
for loops. I mean

if N = 1, the I have

for (int i1 = ....) {
<code>
}

if N = 2, I have

for (int i1 = ...)
for (int i2 = ...) {
<code>
}

....

if N = n I must have

for (int i1 = ...)
....
for (int in = ...) {
<code>
}


Is there a way to program this disregarding the value of N?

thanks

 

[Walter Roberson]

Hi, Somehow I can't figure this one out.

I have an integer variable N and this N reflects the amount of nested
for loops. I mean

if N = n I must have

for (int i1 = ...)
...
for (int in = ...) {
<code>
}

Is there a way to program this disregarding the value of N?

int *indices = malloc(N * sizeof(int));

then indices[N-1] corresponds to iN

 

[jacob navia]

Hi, Somehow I can't figure this one out.

I have an integer variable N and this N reflects the amount of nested
for loops. I mean

if N = 1, the I have

for (int i1 = ....) {
<code>
}

if N = 2, I have

for (int i1 = ...)
for (int i2 = ...) {
<code>
}

...

if N = n I must have

for (int i1 = ...)
...
for (int in = ...) {
<code>
}


Is there a way to program this disregarding the value of N?

thanks

It depends on the value of the upper bound and lower bound obviously
In case all are the same, (upper-lower)^n will do.

For instance:
for (i=0; i<10;i++)
for (j=0; j<10;j++)

<code>

is the same as

for (i=0; i<100; i++)

<code>

100=(10 - 0) ^ 2

jacob

 

[Lew Pitcher]

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Hi, Somehow I can't figure this one out.

I have an integer variable N and this N reflects the amount of nested
for loops. I mean

if N = 1, the I have

for (int i1 = ....) {
<code>
}

if N = 2, I have

for (int i1 = ...)
for (int i2 = ...) {
<code>
}

...

if N = n I must have

for (int i1 = ...)
...
for (int in = ...) {
<code>
}


Is there a way to program this disregarding the value of N?

You /could/ do it with recursion.

i.e.

void function(level)
{
int counter;

for (counter = 1; counter < 10; ++counter)
{
if (level == 0)
{
/* do usefull work */
}
else function(level-1);
}
}



- --
Lew Pitcher

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