How can I store and/or print() a number that is larger than 4 294 967 295 in C? is int64_t or u_int64_t what I need ? if, so how can I printf it to stdout? (2 Replies)
I am working on a re-engineering project. Original Code is written in C. In the C code some "forms" are being called. Each form is in a separate file and files are tagged "int" or "int.lst" like f00.int, f00.int.lst
Can some body through some light on what are these files and what is the... (2 Replies)
Hi,
Is there any way to calculate the size of a built in data type without using 'sizeof' operator? I also don't have the option to read it from std .h file.
regards
Apoorva Kumar (10 Replies)
hello everybody!
I want to create a file with permissions for read, write, and execute to everybody using C, so I write this code:
#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
int main(){
int fileDescriptor;
fileDescriptor =... (2 Replies)
hello guys i m new to shell scripting and can't find out why this structure is not right
I m guessing this happens because $LINESUM is a string . so how can i do this ?
i want my script to do so many loops as the number of the lines of one custom file.
#!/bin/bash
echo give me path name... (5 Replies)
int air_date='20100103'; //2010 - Jan - 03
/* My goal here is to subtract a day. */
int day = air_date % 100; //?????? Is this right?
//Are there any functions time/date for this type of date format?
:cool: (7 Replies)
A simple arithmetic example: 1680 / 1.12 = 1500
My C code result is 1499, here is the code:
#include <stdio.h>
main(int argc, char *argv)
{
int t = 1680;
double adj = 1.12;
int ires = t / adj;
double fres = t / adj;
... (8 Replies)
Hello Everyone,
I am new to awk and trying my hand with the diff codes and came across the below code today. It would be great if any of the Guru's help me to understand.
awk '{filename = "sample_file" int((NR-1)/34) ".DAT"; print >> filename}' sample_file.DAT
34 is the no of lines each... (7 Replies)
Discussion started by: saratha14
7 Replies
LEARN ABOUT DEBIAN
ubuntu-upload-permission
ubuntu-upload-permission(1) General Commands Manual ubuntu-upload-permission(1)NAME
ubuntu-upload-permission - Query upload rights and (optionally) list the people and teams with upload rights for a package
SYNOPSIS
ubuntu-upload-permission [options] package
DESCRIPTION
ubuntu-upload-permission checks if the user has upload permissions for package. If the --list-uploaders option is provided, all the people
and teams that do have upload rights for package will be listed.
OPTIONS -r RELEASE, --release=RELEASE
Query permissions in RELEASE. Default: current development release.
-a, --list-uploaders
List all the people and teams who have upload rights for package.
-t, --list-team-members
List all the members of every team with rights. (Implies --list-uploaders)
-h, --help
Display a help message and exit
EXIT STATUS
0 You have the necessary upload rights.
1 You don't have the necessary upload rights.
2 There was an error.
AUTHORS
ubuntu-upload-permission and this manpage were written by Stefano Rivera <stefanor@ubuntu.com>.
Both are released under the terms of the ISC License.
ubuntu-dev-tools November 2011 ubuntu-upload-permission(1)