Count in octal: Difference between revisions

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The task is to produce a sequential count in octal, starting at zero, and using an increment of a one for each consecutive number. Each number should appear on a single line, and the program should count until terminated, or until the maximum value that can be held within the system registers is reached (for a 32 bit system using unsigned registers, this value is 37777777777 octal).

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=={{header|AWK}}==

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The awk extraction and reporting language uses the underlying C library to provide support for the printf command. This enables us to use that function to output the counter value as octal:

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<lang awk>BEGIN {

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for (l = 1; l < 2147483647; l++) {

printf("%o\n", l);

}

}</lang>

=={{header|C++}}==

This prevents an infinite loop by counting until the counter overflows and produces a 0 again. This could also be done with a for or while loop, but you'd have to print 0 (or the last number) outside the loop.

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<lang cpp>#include <iostream>

#include <iomanip>

using namespace std;

int main()

{

unsigned i = 0;

do

{

cout << setbase(8) << i << endl;

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++i;

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} while(i != 0);

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}</lang>

=={{header|ALGOL 68}}==

{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny].}}

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8r00000000021

</pre>

⚫

=={{header|AWK}}==

⚫

The awk extraction and reporting language uses the underlying C library to provide support for the printf command. This enables us to use that function to output the counter value as octal:

⚫

<lang awk>BEGIN {

⚫

for (l = 1; l < 2147483647; l++) {

printf("%o\n", l);

}

}</lang>

=={{header|C++}}==

This prevents an infinite loop by counting until the counter overflows and produces a 0 again. This could also be done with a for or while loop, but you'd have to print 0 (or the last number) outside the loop.

⚫

<lang cpp>#include <iostream>

#include <iomanip>

using namespace std;

int main()

{

unsigned i = 0;