I'm trying to use command line arguments and have it read into an array however the complier is giving me this error:
error: invalid initializer
I know I need to use int main(int argc, char *argv[]) to use command line arguments and then I have it set like this to read into the array:
int arr[] = atoi(argv[1]);
Not sure what I am missing as this always worked for me in the past. Any ideas?
If what you are passing through the command line a series of number and you want
to have them in an array, then you can do this:
int main(int argc, char **argv)
{
if(argc < 2)
{
fprintf(stderr, "not enough arguments\n");
return 1;
}
int arr[argc-1];
for(size_t i = 0; i < argc - 1; ++i)
arr[i] = atoi(argv[i+1]);
...
return;
}
The reason why the conversion starts at argv[i+1] and not argv[0] is because
argv[0] has always the string that contains the file name of the executed
binary (more precisely the was you've passed the command in the shell), so
argc is always at least 1. That's why the dimension of arr is argc-1,
because for n arguments, argc will be n+1.
Also be aware that atoi does a poor job when it encounters an error, if the
string is not an integer, then atoi will return 0 and you have no idea whether
this is a legit number of an error in the conversion. Using strtol is a
far better alternative:
int main(int argc, char **argv)
{
if(argc < 2)
{
fprintf(stderr, "not enough arguments\n");
return 1;
}
int arr[argc-1];
char *endptr;
for(size_t i = 0; i < argc - 1; ++i)
{
arr[i] = strtol(argv[i+1], &endptr, 0);
if(*endptr != 0)
{
fprintf(stderr, "The argument #%d is not a number\n", i+1);
return 1; // error
}
}
...
return;
}
This would give you a far better result, because it reacts to error from the
user.
Related
I'm trying to check argument count, which should return 1 if there is no argument or more than 2 arguments in command line:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(int argc, char *argv[])
{
char *key = argv[1]; //taking the key from CL
int n = strlen(key); //length of CL arg string
if (argc != 2) //checks the amount of CL args
{
printf("Usage: ./substitution key\n");
return 1;
}
for (int i = 0; i < n; i++) //validating the key array
{
if (n != 26)
{
printf("Key must contain 26 characters.\n");
return 1;
}
else if (!isalpha(key[i]))
{
printf("Key must contain only alphabetic character.\n");
return 1;
}
for (int j = i + 1; j < n; j++)
{
if (key[i] == key[j])
{
printf("Key must not contain repeated characters.\n");
return 1;
}
}
}
}
When i'm running without an argument it throws an error:
runtime error: null pointer passed as argument 1, which is declared to never be null
Segmentation fault (core dumped).
If i'm commenting key validating part, it runs.
I'm new in C programming and googled it a lot.
Please, help me figure out why?
You first access argv[1], which may not exist, then you check argc to see if it exists. This is the wrong order. First check argc, then access argv[1]:
int main(int argc, char *argv[])
{
if (argc != 2) //checks the amount of CL args
{
printf("Usage: ./substitution key\n");
return 1;
}
char *key = argv[1]; //taking the key from CL
int n = strlen(key); //length of CL arg string
// remaining code...
#include <stdio.h>
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "usage: one file only", argv[0]);
return (1);
} else {
for(int i = 0; i < argc; i++) {
putchar(argv[i]);
}
}
}
Say I want to print the input it takes in, for instance
$ gcc -Wall fileabove.c
$ ./a.out abcdefghijlmn
abcdefghijlmn
Basically just prints out whatever text I put into it.
putchar(argv[i]);
is incorrect, because putchar expects a single character (type char), you are passing a pointer (char*). It should be
puts(argv[i]);
So the correct code:
#include <stdio.h>
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "usage: one file only\n", argv[0]);
return 1;
}
// no need for the else
// you exit program anyway if argc != 2
// makes code more readable
puts(argv[1]);
return 0;
}
If you want to print character by character:
#include <stdio.h>
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "usage: one file only\n", argv[0]);
return (1);
}
for(size_t i = 0; argv[1][i] != '\0'; ++i)
putchar(argv[1][i]);
putchar('\n');
return 0;
}
edit changed puts("") to putchar('\n') and removed strlen as Jonathan Leffler mentioned in the comments.
The argument array is always in the format of first index being the called program name and the subsequent index values being the parameters in order. Because you only want one argument to process, I have simplified your code as follows:
#include <stdio.h>
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "%s usage: one file only\n", argv[0]);
return 1;
} else {
printf("%s\n",argv[1]); //print only parameter and new line
}
return 0;
}
When the user doesn't specify one parameter, then the error pops up including the program name. When the parameter is specified, it is stored in the second index of the array (index 1) and in my code, I simply printed it out using printf.
I want to send the size of an 2D array and filename as command line arguments in C.
int main(int argc,char *argv[])
I know this works for single arguments, but how do I take two arguments?
argc tells you the number of arguments.
argv is an array of char pointers to c-style strings.
So you can simply print all arguments by:
int main(int argc, char *argv[])
{
int i;
for (i=0; i<argc; ++i)
{
printf("%s\n", argv[i]);
}
}
You can use atoi to convert (the initial portion of) a string to an integer.
So you can do something like:
int main(int argc,char *argv[])
{
char filename[100];
int size = 0;
int i;
if (argc < 3)
{
printf("Too few arguments\n");
return 0;
}
if (strlen(argv[1]) >= 100)
{
printf("File name too long\n");
return 0;
}
strcpy(filename, argv[1]);
size = atoi(argv[2]);
if (size <= 0)
{
printf("Invalid size\n");
return 0;
}
....
....
return 0;
}
Note that it is usually not necessary to copy file name arguments to another variable unless you are going to modify the value in some way.
So, I am trying to use argc and argv in Caesars cipher in order to execute the program with just [./ key ;string] (e.g. ./ 13 Caesar). I have tried in lots of ways, although I must admit I am really lost here. I was thinking I should just use main(void) and ask the input with fgets, but I still have some curiosity in: How could I make it work with “int main(int argc, char *argv[])?”. Any clues you can give me?
Thank you for your help. Here is the code with the current outputs:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int main(int argc, char *argv[])
{
int key;
int result;
char str[60];
int k = 0;
printf("argc =%d\n", argc);
printf("argv =%d\n", argv);
printf("key =%d\n", key);
if (argc != 2)
{
printf("You didn't enter a key.\n");
return 1;
}
else
{
int k = key % 26;
printf("k =%d\n", k);
if (k == 0)
{
printf("Invalid key.\n");
return 1;
}
}
}
Output:
$ ./ceasar 13
argc =2
argv =-13216
key =0
k =0
Invalid key.
Edit: Tentative Answer
int main(int argc, char* argv[])
{
int i;
int key = atoi(argv[1]);
int result;
char str[60];
for (i = 0; i < argc; i++)
{
printf("argv[%d] = %s\n", i, argv[i]);
}
if (argc != 2)
{
printf("You didn't enter a key.\n");
return 1;
}
else
{
if (key == 0)
{
printf("Invalid key.\n");
return 1;
}
}
}
Printing argv with the "%d" specifier is undefined behavior, read printf()'s manual and use "%p".
On the other hand, if you want to print the string you should access the appropriate element with array notation. For example
printf("First Argument: %s\n", argv[1]);
this applies to all arguments, noting that argv[0] is the name of the executable as invoked in the command line.
You should also be careful before accessing argv[1] to check that argc > 2, and always check the current argument + !.
I'm trying to write a program that finds the largest and smallest of 10 numbers.
To use my program, you must use the command line argument -l then numbers to determine largest number, the same for the command -s for smallest numbers.
However, when I don't enter a command at all, and just try to run the program, I receive a segmentation fault. Not sure where I went wrong.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char* argv[])
{
int i;
int min,max,num;
char *argv1 = argv[1];
char *small = "-s";
char *large = "-l";
min=max=0;
if (0==strcmp(argv1, small))
{
for (i=2; i<argc; i++)
{
num=atoi(argv[i]);
if(i==2)
{
min=num;
}
else
{
if(min>num)min=num;
}
}
printf("The smallest number is %d\n",min);
}
else if (0==strcmp(argv1, large))
{
for (i=2; i<argc; i++)
{
num=atoi(argv[i]);
if(i==2)
{
max=num;
}
else
{
if(max<num)max=num;
}
}
printf("The largest number is %d\n",max);
}
else
{
printf("Invalid option");
}
return 0;
}
Check the number of arguments before accessing to arguments.
int main(int argc, char* argv[])
{
int i;
int min,max,num;
char *argv1 = argv[1];
char *small = "-s";
char *large = "-l";
/* add from here */
if(argc < 2)
{
fprintf(stderr, "Usage: %s command numbers...\n", argc > 0 ? argv[0] : "");
return 1;
}
/* add until here */
min=max=0;
You're setting char *argv1 = argv[1]; without first checking argc to see how many arguments were passed. This will cause a segfault when you later do if (0==strcmp(argv1, small)) because argv1 isn't pointing at a string like you are expecting it to.
To fix it, just check argc before you start comparing argv1 with anything:
if (argc == 1)
{
printf("Error: -s or -l required\n");
exit(1);
}