How to use command line arguments in c (OS: Windows)? - c

I'm unable to make a program that asks the user to input three arguments on the command line:
1) an operator (+, -, *, /);
2) an integer (n);
3) another integer (m).
The program should thus act as a basic calculator producing the output in this format: .
e.g.
operator='+'
n=5
m=6
output: 5+6
= 11

If you want to take the argument from the command line while the user executes the program you can use the argv vector to fetch the values
int main(int argc, char** argv){
}
So that if you execute your program as follows,
./prog + 1 2
argv will contain the follwing,
argv[0] = 'prog',
argv[1] = '+',
argv[2] = '1',
argv[3] = '2',
So that you can fetch each value from argv and implement your logic.
Read this tutorial for better understanding.

If you are interested in giving arguments while executing the file, i.e. ./executable_name arg1 arg2 arg3 then use
int main(int argc, char *argv[])
to get argument from command line. argc gives the count of argument(s) passed including the executable name and argv is the argument array preserving the order of the input.
You can then compile and run code in command prompt:
cl filename.c
executablename arg1 arg2 arg3

Related

Passing file contents to a c executable doesn't seem to work

I have this c executable called testFile file containing this code :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]){
printf("Number of arguments : %d\n Arguments : ", argc);
for(int i = 0; i<argc-1;i++){
printf("%s\t", argv[i]);
}
}
and along with it a file called test1 containing just the number 1 (echo 1 > test1)
When I call this line on the command line (zsh) :
./test < test1
the output I get is this :
Number of arguments : 1
Arguments : ./testFile
Shouldn't this show 2 arguments ? Along with the character 1 ? I want to find a way to make this 1 appear in the arguments list, is it possible ? Or is it just the way my shell handles arguments passed like that ? (I find it weird as cat < test1 prints 1)
You're conflating standard input with command arguments.
main's argc and argv are used for passing command line arguments.
here, for example, the shell invokes echo with 1 command line argument (the 1 character), and with its standard output attached to a newly opened and truncated file test.
echo 1 > test1
here, the shell running test with 0 arguments and its standard input attached to a newly opened test1 file.
./test < test1
If you want to turn the contents of ./test1 into command line parameters for test, you can do it with xargs.
xargs test < test1
unrelated to your question:
for(int i = 0; i<argc+1;i++){
The condition for that should be i<argc. Like all arrays in C and just about every other language, the minimum valid index of argv is 0 and the maximum valid index is 1 less than its length. As commenters pointed out, argv[argc] is required to be NULL, so technically, argc is one less than the length of argv.
If you want the contents of test1 to be available in argv, you can do:
./test $(cat test1)
Note that this is very different than:
./test "$(cat test1)"
which is also different from:
./test '$(cat test1)'. # Does not do at all what you want!
The first option will present each "word" of the file as a distinct argument, while the second will present the contents of the file as a single argument. The single quotes version doesn't look inside the file at all, and is merely included here to give you something to experiment with.
argv[0] is always the name of the executable itself. With your way of invoking the command, you did not pass any command at all. You would have to invoke it from zsh as ./testFile test1 if you want your program to see the name of the data file, or ./testFile $(<test1) if you want it to see the content of the data file.

Provide input parameters to C code from Linux shell

The concept of my code is like:
#include <stdio.h>
int main(int argc, char *argv[])
{
int num;
FILE *fp;
getint("num",&num); /* This line is pseudo-code. The first argument is key for argument, the second is the variable storing the input value */
fp = inputfile("input"); /* This line is pseudo-code. The argument is key for argument, fp stores the return file pointer */
...
...
exit(0);
}
Usually, after compiling the code and generating the executable main, in the command line we write this to run the code:
./main num=1 input="data.bin"
However, if there's too many arguments, type in the command line each time we run the code is not convenient. So I'm thinking about writing the arguments and run in Linux shell. At first I wrote this:
#! /bin/sh
num = 1
input="data.bin"
./main $(num) $(input)
But error returns:
bash: adj: command not found
bash: input: command not found
bash: adj: command not found
bash: input: command not found
Can anybody help to see and fix it.
There are three main problems with your code:
You can't use spaces around the = when assigning values
You have to use ${var} and not $(var) when expanding values.
The way your code is written, you are passing the string 1 instead of your required string num=1 as the parameter.
Use an array instead:
#!/bin/bash
parameters=(
num=1
input="data.bin"
)
./main "${parameters[#]}"
num=1 is here just an array element string with an equals sign in it, and is not related to shell variable assignments.

run linux command using execlp with more than one argument as string in c

I am trying to run ls using system calls in C with more than one argument, for example -l -a. The arguments and their number is changing depending on the user input. The input is concatenated "-l" + "-a" == "-l -a". The code I'm using is:
execlp("ls","ls",arguments,NULL) //arguments = "-l -a"
The user input is from Terminal:
-l
-a
if you want to executes more than one argument , then you should use execvp() instead of execlp.
#include<stdio.h>
#include <unistd.h>
int main(int argc,char *argv[])
{
execvp(argv[1],argv+1);// argv+1 means whatever arguments after argv[1] it will take & executes it
return 0;
}
for e.g your input like that
xyz#xyz-PC:~$ ./a.out ps -el
I hope it helps.

How do I run a python script and pass arguments to it in C

I have a python script script.py which takes command line params
I want to make a wrapper in C so I can call the script.py using ./script args
So far I have this in my script.c file
#include<stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]){
system("python3.4 script.py");
return 0;
}
How do I modify the script so I can do ./script arg1 arg2 and the C code executes system("python3.4 script.py arg1 arg2");
I don't have experience in C. Above code is from googling
Using system() is needlessly complicated in this case, as it effectively passes the given command string to (forked) sh -c <command>. This means that you'd have to handle possible quoting of arguments etc. when forming the command string:
% sh -c 'ls asdf asdf'
ls: cannot access 'asdf': No such file or directory
ls: cannot access 'asdf': No such file or directory
% sh -c 'ls "asdf asdf"'
ls: cannot access 'asdf asdf': No such file or directory
Note the difference between the unquoted and quoted versions.
I'd suggest using execve(), if executing the python command is the only purpose of your C program, as the exec family of functions do not return on success. It takes an array of const pointers to char as the new argv, which makes handling arguments easier:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#define PYTHON "/usr/bin/python3"
#define SCRIPT "script.py"
int
main(int argc, char *argv[])
{
/* Reserve enough space for "python3", "script.py", argv[1..] copies
* and a terminating NULL, 1 + 1 + (argc - 1) + 1 */
int newargvsize = argc + 2;
/* VLA could be used here as well. */
char **newargv = malloc(newargvsize * sizeof(*newargv));
char *newenv[] = { NULL };
newargv[0] = PYTHON;
newargv[1] = SCRIPT;
/* execve requires a NULL terminated argv */
newargv[newargvsize - 1] = NULL;
/* Copy over argv[1..] */
memcpy(&newargv[2], &argv[1], (argc - 1) * sizeof(*newargv));
/* execve does not return on success */
execve(PYTHON, newargv, newenv);
perror("execve");
exit(EXIT_FAILURE);
}
As pointed out by others, you should use the official APIs for this, if at all possible.
You can generate your command as a string. You just need to loop through argv[] to append each parameters given to the C program at then end of your command string. Then you can use your command string as the argument for the system() function.

Handling command line flags in C/C++

I am looking for a very simple explanation/tutorial on what flags are. I understand that flags work indicate a command what to do. For example:
rm -Rf test
I know that the rm command will remove the test folder and that the -Rf flags will force the command to erase not just the folder but the files in it.
But, where are the flags read/compiled??? What handles the flags? Can I, for example, write my own C/C++ program and designate different flags so that the program does different things? I hope I am asking the right questions. If not, please let me know.
At the C level, command line arguments to a program appear in the parameters to the main function. For instance, if you compile this program:
#include <stdio.h>
int main(int argc, char **argv)
{
int i;
for (i = 0; i < argc; i++)
printf("argv[%d] = %s\n", i, argv[i]);
return 0;
}
and invoke it with the same arguments as your example 'rm' command, you get this:
$ ./a.out -Rf test
argv[0] = ./a.out
argv[1] = -Rf
argv[2] = test
As you can see, the first entry in argv is the name of the program itself, and the rest of the array entries are the command line arguments.
The operating system does not care at all what the arguments are; it is up to your program to interpret them. However, there are conventions for how they work, of which the following are the most important:
Arguments are divided into options and non-options. Options start with a dash, non-options don't.
Options, as the name implies, are supposed to be optional. If your program requires some command-line arguments to do anything at all useful, those arguments should be non-options (i.e. they should not start with a dash).
Options can be further divided into short options, which are a single dash followed by a single letter (-r, -f), and long options, which are two dashes followed by one or more dash-separated words (--recursive, --frobnicate-the-gourds). Short options can be glommed together into one argument (-rf) as long as none of them takes arguments (see below).
Options may themselves take arguments.
The argument to a short option -x is either the remainder of the argv entry, or if there is no further text in that entry, the very next argv entry whether or not it starts with a dash.
The argument to a long option is set off with an equals sign: --output=outputfile.txt.
If at all possible, the relative ordering of distinct options (with their arguments) should have no observable effect.
The special option -- means "do not treat anything after this point on the command line as an option, even if it looks like one." This is so, for instance, you can remove a file named '-f' by typing rm -- -f.
The special option - means "read standard input".
There are a number of short option letters reserved by convention: the most important are
-v = be verbose
-q = be quiet
-h = print some help text
-o file = output to file
-f = force (don't prompt for confirmation of dangerous actions, just do them)
There are a bunch of libraries for helping you parse command line arguments. The most portable, but also the most limited, of these is getopt, which is built into the C library on most systems nowadays. I recommend you read all of the documentation for GNU argp even if you don't want to use that particular one, because it'll further educate you in the conventions.
It's also worth mentioning that wildcard expansion (rm -rf *) is done before your program is ever invoked. If you ran the above sample program as ./a.out * in a directory containing only the binary and its source code you would get
argv[0] = ./a.out
argv[1] = a.out
argv[2] = test.c
This simple program should demonstrate the arguments passed to the program (including the program name itself.)
Parsing, interpreting and using those arguments is up to the programmer (you), although there are libraries available to help:
int main(int argc, char* argv[])
{
int i;
for(i=0; i<argc; ++i)
{ printf("Argument %d : %s\n", i, argv[i]);
}
return 0;
}
If you compile this program into a.out, and run it as:
prompt$> ./a.out ParamOne ParamTwo -rf x.c
You should see output:
Argument 0 : a.out
Argument 1 : ParamOne
Argument 2 : ParamTwo
Argument 3 : -rf
Argument 4 : x.c
Actually you can write your own C++ programm which accepts commandline parameters like this:
int main(int argc, char* argv[]){}
The variable argc will contain the number of parameters, while the char* will contain the parameters itself.
You can dispatch the parameters like this:
for (int i = 1; i < argc; i++)
{
if (i + 1 != argc)
{
if (strcmp(argv[i], "-filename") == 0) // This is your parameter name
{
char* filename = argv[i + 1]; // The next value in the array is your value
i++; // Move to the next flag
}
}
}
In your own C program you can process command line options in any way you see fit.
Command line parameters in C come in the parameters of the main(int argc, char *argv[]) method as strings.
And if you'd like to process command line parameters in a way similar to most UNIX commands, the function you're probably looking for is getopt()
Good luck!
The easiest thing is to write your main() like so:
int main(int argc, char* argv[]) { ...
Then inside that main you decide what happens to the command line arguments or "flags". You find them in argv and their number is argc.
flags are arguments passed into the main entry point of the program. For example, in a C++ program you can have
int main(int arc, char* argv[]){
return 0;
}
your arc is the # of arguments passed in, and the pointer gives u the list of actual arguments. so for
rm -Rf test
argc would be 3, and the argv array would contain your arguments. Notice argc >= 1 because the program name itself counts (rm). -RF is your 2nd parameter and test is your third.
So whenever you are typing commands in unix, you essentially are executing programs and passing them parameters that they operate on.
If you are really REALLY interested in the unix OS, you should look up forks and how they work. This can get pretty confusing to a newcomer though, so only if you are really interested in OS and how programs are executed.
GNU libc, which is very likely available on your system, has a library for this called getopt that can be used to parse the options in a sensible fashion. There are examples to get you started in the documentation linked below.
http://www.gnu.org/software/libc/manual/html_node/Getopt.html#Getopt

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