Why aren't there any passed arguments to argv[]? - c

So, I noticed that my argc is always 1 as I will always get the message Error: missing command line arguments!, but as stated in code I am using argv[1] and argv[2] to read the files names.
Shouldn't automatically argc be 3 in this case, and to be able to pass that error?
Notes:
If I am not using if (argc < 2) statement, after I enter the name for the output file my program crushes.
For the input file I already have this one created in the project folder, so I just enter the name of that file.
This is the code:
#include <stdio.h>
#include <stdlib.h>
FILE *IN, *OUT;
int main(int argc, char* argv[])
{
if (argc < 2)
{
printf("Error: missing command line arguments!\n");
return 1;
}
printf("Enter the name of the file with the input data: ");
scanf("%s", argv[1]);
printf("\nEnter the name of the file for the output data: ");
scanf("%s", argv[2]);
IN = fopen(argv[1], "r");
OUT = fopen(argv[2], "w");
fclose(IN);
fclose(OUT);
return 0;
}

You're complete mis-understanding the purpose of argc and argv. They are supposed to receive the supplied command line argument (program parameters) before the program startup, not supposed to hold the scannned input at runtime.
Quoting C11, chapter §5.1.2.2.1,
If they are declared, the parameters to the main function shall obey the following
constraints:
— The value of argc shall be nonnegative.
— argv[argc] shall be a null pointer.
— If the value of argc is greater than zero, the array members argv[0] through
argv[argc-1] inclusive shall contain pointers to strings, which are given
implementation-defined values by the host environment prior to program startup. The
intent is to supply to the program information determined prior to program startup
from elsewhere in the hosted environment.
and
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.
To elaborate, when a program is invoked like
./a.out three total arguments
then, in your program
argc will be 4
argv[0] will be ./a.out
argv[1] will be three
argv[2] will be total
argv[3] will be arguments
argv[4] will be NULL (see the property for argv[argc] above)
(To generalize, argv[1] - argv[argc-1] will hold the supplied arguments, argv[argc] will be NULL)
You don't need to explicitly scan the input, those values will be populated from the host environment.
On the other hand, you cannot just invoke the program like
./a.out
and exptect argc to be 3 (or any other value, other than 1, for that matter) and argv[1] - argv[n-1] to be valid because at compile-time the invokes program has no way to know that you plan to supply some values at runtime. It's not clairvoyant.

Related

Segmentation fault when file is not found even if I try to create it in C

I'm writing a code that appends text to a file.
It uses fopen in the beginning of WriteToFile so even if the file does not exist, it creates it.
But, what happens when I enter no file at all? no arguments at all? Just ./a.out?
Segmentation fault.
I don't know why, it seems to me I did everything fine to avoid any problems.
int main(int argc, char **argv)
{
char file_name[30] = "file.txt";
printf("%s",file_name); /* for debugging : doesn't print it */
if (0 != argc)
{
strcpy(file_name, argv[1]);
}
WriteToFile(file_name);
}
OR
(in case I can't really put a string literal into char array):
char file_name[30];
if (0 == argc)
{
strcpy(file_name, "file.txt");
}
else
{
strcpy(file_name, argv[1]);
}
For both of the cases i'm getting
Segmentation fault (core dumped)
if (0 != argc)
The argc value is normally(a) the count of arguments including the program name. Hence, running ./a.out will have an argc of one rather than zero. And, since argv[argc] is usually NULL, dereferencing it is not going to end well :-)
If you want to ensure another argument is available, you can use:
if (argc > 1) { // Have both argv[0] AND argv[1].
nowSafeToUse(argv[1]);
}
In more detail, the C11 standard states:
If they are declared, the parameters to the main function shall obey the following constraints:
The value of argc shall be nonnegative.
argv[argc] shall be a null pointer.
If the value of argc is greater than zero, the array members argv[0] through argv[argc-1] inclusive shall contain pointers to strings, which are given implementation-defined values by the host environment prior to program startup. The intent is to supply to the program information determined prior to program startup from elsewhere in the hosted environment. If the host environment is not capable of supplying strings with letters in both uppercase and lowercase, the implementation shall ensure that the strings are received in lowercase.
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.
The parameters argc and argv and the strings pointed to by the argv array shall be modifiable by the program, and retain their last-stored values between program startup and program termination.
As an aside regarding this line:
printf("%s",file_name); /* for debugging : doesn't print it */
If this is not printing, it's probably because standard output is line buffered (default if it's determined to be an interactive device, otherwise fully buffered).
By not outputting a \n at the end, the characters are probably still sitting in a buffer somewhere, ready to be written. The crash will probably kill off the process without flushing the buffer. So a simple solution may be just to use one of:
printf("%s",file_name);
puts(file_name);
As another aside, you're going to get into trouble if the filename you enter is larger than 29 characters since it will overflow file_name, allowing for the \0 at the end as well.
A better approach may be just to use either your default string or argv[1] directly (without copying), something like:
int main(int argc, char **argv) {
char *file_name = (argv > 1)
? argv[1]
: "file.txt";
printf("%s\n", file_name); // for debugging : probably does print it :-)
WriteToFile(file_name);
}
(a) Not required by the standard since it allows for implementation-specific differences, but it's the usual case. Specifically, the phrase:
... which are given implementation-defined values by the host environment prior to program startup. The intent is to supply to the program information determined prior to program startup from elsewhere in the hosted environment.
pretty much means it can do whatever it wants :-)
A related answer (though a non-duplicate question) can be found here.
When argc is 1, argv is treated as an array of 2 pointers to char. The first is a valid pointer (typically to the program name), the second is NULL.
argv[1] accesses the second element of argv. With no arguments supplied, argc will be 1 and argv[1] will be NULL. You're therefore dereferencing a NULL pointer.
The condition 0 != argc should instead be argc >= 2 or argc > 1, and the condition 0 == argc should be argc < 2 or argc <= 1.
You have to account for the program's name so you should write if (argc >= 2) instead. See the man
The value of the argc argument is the number of command line arguments. The argv argument is a vector of C strings; its elements are the individual command line argument strings. The file name of the program being run is also included in the vector as the first element; the value of argc counts this element. A null pointer always follows the last element: argv[argc] is this null pointer.
argv[1] means index 1 but in C you start at index 0.
Fixed code
int main(int argc, char **argv)
{
char file_name[30] = "file.txt";
printf("%s",file_name); /* for debugging : doesn't print it */
if (argc >= 2)
strcpy(file_name, argv[1]);
else
return 1; // exit program with error code.
WriteToFile(file_name);
return 0; // exit with success
}

What Does Input Arguments is _

I wanna know why is result 1 for no arguments.
int main(int argcount, char *arglist[]) {
int i;
printf("Number of arguments %d\n",argcount);
printf("Arguments list:\n");
for (int i=0;i<argcount;i++)
printf("%s\n",arglist[i]);
return EXIT_SUCCESS;
}
From the standard (C11), noting especially the bold bit:
If they are declared, the parameters to the main function shall obey the following
constraints:
The value of argc shall be nonnegative.
argv[argc] shall be a null pointer.
If the value of argc is greater than zero, the array members argv[0] through
argv[argc-1] inclusive shall contain pointers to strings, which are given
implementation-defined values by the host environment prior to program startup. The
intent is to supply to the program information determined prior to program startup
from elsewhere in the hosted environment. If the host environment is not capable of
supplying strings with letters in both uppercase and lowercase, the implementation
shall ensure that the strings are received in lowercase.
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.
The parameters argc and argv and the strings pointed to by the argv array shall
be modifiable by the program, and retain their last-stored values between program
startup and program termination.
In other words, argc includes the argument representing the program name - the actual parameters to the program start at argv[1]. That becomes evident from the program output as per the following transcript, where the first argument is the program name:
pax> cat testprog.c
#include <stdio.h>
int main(int argc, char *argv[]) {
printf("Argument count: %d\n", argc);
printf("Arguments:\n");
for (int i = 0; i < argc; i++)
printf(" %s\n", argv[i]);
return 0;
}
pax> gcc --std=c11 -o testprog testprog.c && ./testprog 1 2 3
Argument count: 4
Arguments:
./testprog
1
2
3
pax> ./testprog
Argument count: 1
Arguments:
./testprog
./a.out is the first argument so that's why argcount equals to 1.
From the C Standard (5.1.2.2.1 Program startup. p.#2)
— 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.
So the used by you environment supplies the program name in the first argument.

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

One command line argument in 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.

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

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