One command line argument in C - c

write c program that accepts one command line argument (your first name) and prompts the user for user input (your last name), then print ""Welcome to operating systems, "" to the screen.
Can anyone help me with this question? I know its using something like this from the below, but I dunno how to print out the thing? Can anyone help by giving the full answer? Thanks in advance.
int main (int argc, int *argv[])

argc is an integer that represents the number of command line arguments passed in to the program. It is the argument count, hence the name. *argv[] (or **argv depending on developer preference) represents the actual arguments. The proper name for argv is argument vector, which makes sense if you're familiar with that particular data type.
The first argument passed in, argc = 1 is the program's name. Argc is always at least one because argv will always contain at a minimum the name of the program.
To answer your question, you need to pass in a second command-line argument, argc = 2, where argv[1] equals the user's first name. We can accomplish that like this:
int main(int argc, char** argv)
{
// This line will print out how many command line arguments were passed in.
// Remember that it will always be at least one because the name of the program
// counts as an argument.
printf("argc: %d", argc);
// Remember that you want the second argument in argv,
// so you have to call argv[1] because arrays in C
// are 0-index based. Think of them as offsets.
printf("\nWelcome, %s", argv[1]);
return 0;
}
This should get you started. All you need to do now is write the code to read the string from the standard input and output it to the console.

Related

C Commandline Arguments: Need Clarification About Order of Arguments Input And Commandline Arguments In General

This is a sample code from a book. This program prints a given string to repeat given number of times.
#include <stdio.h>
#include <stdlib.h>
void usage(char *program_name)
{
printf("Usage: %s <nessage> <# of times to repeat>\n", program_name);
exit(1);
}
int main(int argc, char *argv[]) {
int i, count;
if(argc < 3)
usage(argv[0]);
count = atoi(argv[2]);
printf("Repeating %d times..\n", count);
for(i=0; i < count; i++)
printf("%3d - %s\n", i, argv[1]);
}
It does what it should do:
kingvon#KingVon:~/Desktop/asm$ ./convert 'Stackoverflow is the best place to ask questions about programming' 6
Repeating 6 times..
0 - Stackoverflow is the best place to ask questions about programming
1 - Stackoverflow is the best place to ask questions about programming
2 - Stackoverflow is the best place to ask questions about programming
3 - Stackoverflow is the best place to ask questions about programming
4 - Stackoverflow is the best place to ask questions about programming
5 - Stackoverflow is the best place to ask questions about programming
kingvon#KingVon:~/Desktop/asm$
Q. Now although main takes two arguments in this specific order: (int argc, char *argv[]), why is that when I ./convert 'string' (number) it works fine but other way around `./convert (number) 'string' does not work?
kingvon#KingVon:~/Desktop/asm$ ./convert 5 'Stackoverflow is the best place to ask questions about programming'
Repeating 0 times..
Q. This line
if(argc < 3) usage(argv[0]);
I have 2 questions about: This line specifies that if the integer argument given is less than 3, the program should output the usage. ./convert 'string' 2 does not print the usage? So what is happening here? Also usage takes char *program_name as an argument(what is meant by char *program_name?) But in line above is given argv[0] as an argument. Why is this and what does this do?
argc is the number of arguments on the command line, not the value of any particular argument. argv contains the actual arguments, which are passed as strings. argv[0] is the command used to invoke the program, argv[1] is the first argument, etc.
When you call the program as
./convert 'Stackoverflow ...' 6
then
argv[0] == "./convert"
argv[1] == "Stackoverflow ..."
argv[2] == "6”
argc == 3
The code assumes that the number is passed in argv[2] and uses the atoi function to convert it from a string representation of an integer to an integer value, which is why the code didn’t behave as expected when you switched the order of the arguments. If you want to be able to switch up the order of the arguments, then your code has to know how to detect which argument is which.
The argc variable is the number of elements in the argv array. And the actual command line arguments will be in the argv array.
The first argument on the command line will always be in argv[1], the second in argv[2], etc.
If you change the order when running the program, the program doesn't know about that, and will think that the string to print will still be in argv[1] and the number in argv[2]. If that's not true the program will fail to work properly.
The argc check only check the number of arguments, not their order.
The name of the program ("./convert" in your question) is always passed as argument zero, i.e. in argv[0].

What is the definition of an "argument in C"

I'm having some trouble reading a line of the code, and understanding what constitutes an argument in the context of this line of code. This is saved in a file called argv0.c
#include <cs50.h>
#include <stdio.h>
int main(int argc, string argv[])
{
if (argc == 2)
{
printf("hello, %s\n", argv[1]);
}
else
{
printf("hello, world\n");
}
}
I compile the code as follows:
make argv0
./argv0
following which I am prompted for an input. Herein lies the issue:
if I type in "Dion Lim" in the terminal, is Dion Lim considered an argument? If so, is it two arguments?
Why is it that if I type in "Dion Lim" in the terminal, I get "Hello, World", but if I type in "Dion" i get "Hello,Dion"
Q1) Yes, they are two arguments.
Q2) Because argc consider the name of executable it's the first parameter. So:
./argv0 Dion Lim // argc == 3
./argv0 Diom // argc == 2
./argv0 // argc == 1
You can get more detail here.
if I type in "Dion Lim" in the terminal, is Dion Lim considered an
argument? If so, is it two arguments?
It depends on how your shell handles it of course, but usually "Dion Lim" would be one argument while Dion Lim (without the quotation marks) would be two arguments. The space delimits the arguments, and with the quotes you can work around that if you want space in your input (sometimes you can also escape the space, like Dion\ Lim).
Why is it that if I type in "Dion Lim" in the terminal, I get "Hello,
World", but if I type in "Dion" i get "Hello,Dion"
The argc parameter tells you how many arguments you have (I think of it as standing for "argument count"). The program's name also counts as an argument, so if you only pass Dion, then argc will be 2 already. If you pass Dion Lim, it will be 3.
To see the number of arguments check the value argc (arguments count). There is always at least one input argument, which is the program name.
So with./argv0 Dion Lim there are three input arguments.
If you are wondering make compiles the program using Makefile so if look in the directory you from which you are running make you will find a file named Makefile containing the compilation instructions.
According to the C Standard (5.1.2.2.1 Program startup)
— If the value of argc is greater than zero, the string pointed to
by argv[0] represents the program name; argv[0][0] shall be the null
character if the program name is not available from the host
environment. If the value of argc is greater than one, the strings
pointed to by argv[1] through argv[argc-1] represent the program
parameters.
So if you "manually" supply the argument Dion then argc will be exactly equal to 2. The first program parameter will be the program name (as it is followed from the quote) and the second program parameter will be the word Dion.
If you will type Dion Lim then the host system considers them as two program parameters and together with the program name argc will be equal to 3.
However if you enclose the input Dion Lim in parentheses like "Dion Lim" then the system will consider the input as one parameter and your program will output
hello Dion Lim

Open file from command line and display what was typed

I'm trying to input only one filename from command line. Then I want to use an If statement to compare the filename to 4 different names I'm expecting to see. If I don't get one of the 4 expected file names then I need to print it back to the user with what was inputted and then exit the program safely.
int main(int argc, char *argv[])
{
....
}
I've been trying a lot of different methods of getting this done, but I just can't figure it out. I was thinking maybe the way I take the input argument is wrong. Any help would be greatly appreciated.
-edit
I just want to be clear I don't want you to be a leech and solve the question I have for me. Just at least point me in the correct direction. I can't figure how to make a for loop work with the filename.
for(argv == "UnexpectedFile.csv"){
printf("this is an unexpected file: %c", argv[1]);
}
You could use strcmp() from string.h to compare strings like
strcmp(argv[1], "unexpectedfile.csv");
It returns 0 when the strings are equal.
If you have the 4 file names in as an array of strings, say expectedFile, do
for(i=0; i<4 && strcmp(argv[1], expectedFile[i])!=0; ++i);
If the value of i is the total number of file names (ie, 4) after this loop, argv[1] is an unexpected file.
Otherwise, value of i would be the index of the file name string in the expectedFile array.
The command line arguments are stored in the 2-dimensional char array argv.
argv[0] would be the name of the executed file. The arguments you give start only from argv[1] onwards.
argc will have the total number of command line arguments including the file name stored in argc. So if there are 'no' arguments, argc would be 1.
In your case the file name is the only argument, so argc would be 2.
You must check if argc is at least 2 before you access argv[1] to prevent the program from accessing argv[1] when it isn't there.
(You do not ask for complete code solution (and do not provide enough of your code for that). So here are the pointers in the right direction you want.)
The comparison you do with a simple pointer == pointer does not really compare the content of the strings. That is what e.g. strcmp () is for, as proposed by #user3629249.
And in order to print out what was given as commandline argument, you should use "%s\n".
And in order to set up a for loop, you will have to do the syntax right: for(init assignment; condition; step operation).
(If you need more help, you will have to provide more details on what behaviour you get and what you do not like about it. Currently your code looks more like compiler errors, which you did not quote, than a problem with the behaviour goal...)

Reading a text file from the command line C

If someone can help me get past this roadblock that would be amazing. I am trying to open a text file called "inputfile.txt" and I can not! Every example I have looked at has worked fine, but when I try to use it, the file returns null and I get a segmentation fault.
Note, this is prior to error checking
#include <stdio.h>
int main(int argc, char *argv[])
{
FILE *inputPtr;
inputPtr = fopen(argv[2], "r");
fclose(inputPtr);
}
I realized that I "goofed" slightly, when I first created the text files on my desktop I titled it "inputfile.txt", but it was saved as "inputfile.txt.txt" as odd as that is!
Well the first big problem is that you do not check the value of argc. This value is set to the number of arguments that were passed to the programm. By default argc is at least 1, because 1 argument is always passed. If you pass any additional arguments, argc must be greater than 1. In your case, i guess it should be 2.
The second problem comes from the first. The numbering of array elements in C starts from 0, so if your programm accepts argc arguments, argv which keeps the arguments will have argc elements, BUT!!! the last element will have an index of argc-1. By default, if no additional arguments were passed only argv[0] exists and it is the name of the programm, it is also always passed, hence argc is always at least 1.
In your case, if argc==2, then argv[2] simply does not exist, only argv[0], and argv[1] exist. And when you pass one argument to the programm, it will be kept in argv[1]. That means that this line inputPtr = fopen(argv[2], "r"); should be changed to this inputPtr = fopen(argv[1], "r");. Also there should be a check of argc at the beginning of the programm. soome thing along the lines of
int main(int argc, char *argv[])
{
if(argc!=2)
{
printf("Wrong number of arguments\n");
return -1;
}
.....
}
before accessing anything beyond argv[0] (which is always available)
the code must check the value of argc to assure the command line parameter actually exists.
Note; in C, array indexs start from 0 through (count of array entries -1).
So if argc is 2, then there is only one command line parameter and:
argv[0] is the executing program name
argv[1] is the first parameter.
argv[2] is a NULL pointer.
so using argv[2] is accessing address 0 and will result in
undefined behaviour
lead to a seg fault event

C-programming board game [duplicate]

This question already has answers here:
How to parse integer command line arguments in C?
(8 answers)
Closed 9 years ago.
I am trying to make a board game with Maximum board width 80 cells and Maximum board height 52 cells, however the user should be able to choose the dimensions by entering them into the command line, so the dimensions could be smaller than the maximum. I am new to programming but i know that command line arguments are passed through the main function in C, I've just been stuck on this and cant seem to find any answers.
Can anyone help
thank you.
The main function takes two arguments, an integer usually called argc and an array of string pointers usually called argv. main is so usually declared as:
int main(int argc, char *argv[])
The argc variable is the number of arguments passed to your program.
The argv variable is contains the actual arguments pass to your program.
The argc variable is at least equal to 1, as the actual program name is always passed as an argument, and is in argv[0].
If you want two arguments, then you first have to make sure that argc is at least equal to 3. The first argument to your program is then stored as a string in argv[1], and the second is in argv[2].
I recommend you experiment with a program such as this:
#include <stdio.h>
int main(int argc, char *argv[])
{
printf("argc = %d\n", argc);
for (int i = 0; i < argc; i++)
printf("argv[%d] = \"%s\"\n", i, argv[i]);
return 0;
}
And finally (and unrelated to your problem but interesting none the less) is the fact that the size of the argv array is actually one larger than most people expect. The size of the argv array is actually argc + 1 and so can be indexed from argv[0] (the program name) to argv[argc]. This last entry (argv[argc]) is always a NULL pointer.

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