I want to write a C program that takes as input an integer, and outputs its square. Here is what I have tried.
However,
./a.out < 3 outputs 32768, and
./a.out < 4 also outputs 32768.
Where am I wrong? Thanks.
#include <stdio.h>
int main(){
int myInt;
scanf("%d", &myInt);
printf("%d\n",myInt*myInt);
}
It looks like what you're trying to do is
echo 4 | ./a.out
the syntax for < is
program < input_file
whereas | is
command_with_output | program
./a.out < 4
This tries to read the file named 4 and use it's content as input to a.out You can do it whichever way, but understand the < operator isn't for inputting the character you type quite literally.
One way to do this would be:
echo "4" > 4
./a.out < 4
I just run your program and its works great.
If you want the program receive an integer input you should use argc , argv as folowed and not use scanf.
*The code for argc argv: *
#include <stdio.h>
#include <stdlib.h>
int main(int argc , char** argv)
{
int myInt;
myInt = atoi(argv[1]);
printf("%d\n",myInt*myInt);
}
atoi - convert char* to integer.
If you want to run the program and then insert an integer, you did it right!
you can read about atoi
To run this program you should comile and run from terminal:
gcc a.c -o a
./a 3
and you will receive:
9
On the right hand side of "<", there should be a file containing the input.
try this thing:
$ echo "3" > foo
$ ./a.out < foo
Read this for more information (Specially section 5.1.2.2):
http://www.tldp.org/LDP/intro-linux/html/chap_05.html
Related
Here is the code.
#include<stdio.h>
int main(int argc,char* argv[])
{
fprintf(stderr,"arg count:%d\n",argc);
return 0;
}
When type the following cmd:
./a.out x 3 >1.txt //case1
I got argc 3. (in this case, a.out x and 3)
But when type the following cmd:
./a.out x 3>1.txt //case2
I got argc 2. It seems 3>1 was ignored.
I expected the argc is 3 like case1.
(Clarification: There is a space between 3 and > in case1)
Why I got 2nd case? And how to solve it?
First of all you have to remember that all redirection is handled by the shell itself, it's not passed on to the program.
The command
./a.out x 3 >1.txt
have three arguments to a.out: The command itself, x and 3.
With
./a.out x 3>1.txt
then 3>1 is all part of the redirection handled by the shell. It will redirect descriptor 3 to the file 1.txt.
I ran the following just for fun, but cannot account for the result. Assume ./test WTF? was run at the command line, and the output produced was
WTF? 6 2. Why is there such a vast discrepancy between the reported value of argc (2 - as expected) and strlen(*argv), which came up as 6. I know an array of strings isn't exactly what strlen is looking for, however, I thought the value it produced (if it produced anything) would be reasonably close to argc. Thoughts?
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main(int argc, char**argv)
{
for (int i = 1; i < argc; i++)
{
for (int j = 0; j < strlen(argv[i]); j++)
{
printf("%c", argv[i][j]);
}
printf(" ");
}
printf("\t%lu\t%d", strlen(*argv), argc);
printf("\n");
}
Nothing wrong here. The argv array contains two strings -
program name as invoked --> "./test"
"WTF?".
So in your example strlen(argv[0]) will print 6 and strlen(argv[1]) will print 4.
For example, if you run gcc -o myprog myprog.c in command line, you will get the followings:
argc
4
argv[0]
gcc
argv[1]
-o
argv[2]
myprog
argv[3]
myprog.c
Therefore, for your case you will get 6 (length of ./test for argv[0]) instead of 4 (length of WTF? argv[1]) as you run ./test WTF? in command line.
Is it possible to write a script to run this code for different values of A;
#include <stdio.h>
#define A 3
int main (){
printf("In this version A = %d\n", A);
return(0);
}
I guess something like for loop?
Is it possible to write a script to run this code for different values of A;
Not as it is because the macro A has a fixed value defined in your code. Instead you can pass the value as an argument:
#include <stdio.h>
int main(int argc, char **argv){
if(argc == 2) {
printf("In this version A = %s\n", argv[1]);
}
return 0;
}
(The code doesn't check if its input is an integer -- which you can test if necessary).
and you can run it via script. For example, compile the above (gcc -Wall -Wextra test.c -o test) using a for loop of bash:
$ for ((i = 0; i < 10; i++)); do ./test $i; done
In this version A = 0
In this version A = 1
In this version A = 2
In this version A = 3
In this version A = 4
In this version A = 5
In this version A = 6
In this version A = 7
In this version A = 8
In this version A = 9
$
No. But you can make A a command line arg:
#include <stdio.h>
int main (int argc, char *argv[]) {
int a;
if (argc != 2 || sscanf(argv[1], "%d", &a) != 1) return 1;
printf("In this version A = %d\n", a);
return 0;
}
Compile to a binary named foo, then
foo 42
will print
In this version A = 42
You can also compile different versions by defining A in the compilation command line. From your original program, remove the #define. Then
gcc -DA=42 foo.c -o foo
./foo
will print the same as above.
DO you need run program repeated from script? why not to make program that accepts arguments from command line?
1)The main() function actually takes arguments, you can compile program once and pass different parameters, as shown in answers above
2) If you need to change some code parameters from make script, I'd say, create separate header that would contain defines and write script that would echo into that file (> for start, >> to continue writing).
3) Alternative way you can call you compiler with flag that would be equal to #define macro-command. For gcc it's -D, for example -DA=3 instead of #define A 3.
Most programs use makefile to be compiled. For that case you can script make file to use 2) or 3) Former is preferable because you do not need to pass that argument to all compilation targets, reducing time or re-compiling. There are tools for more advanced manipulations, like autoconf.
I have to program a shell in C and need to handle globing in it and I am only allowed to use the function glob. But when I try to use it, it only gives me one result back.
#include <glob.h>
#include <stdlib.h>
int main(int ac, char **av)
{
glob_t s;
unsigned long i = -1;
if (glob(av[1], 0, NULL, &s))
printf("ERROR !\n");
else
while (++i < s.gl_pathc)
printf("%s\n", s.gl_pathv[i]);
return (0);
}
I run this code in a folder where there is two C files : replace_glob.c and test.c
And when I run this code :
$ ./a.out *.c
replace_glob.c
$
I dont understand why and I would really appreciate your help
In the command line
./a.out *.c
the shell expands the glob pattern, so your program sees
{"./a.out", "replace_glob.c", "test.c", NULL}
as its argv. You need to quote the pattern for the program to see it:
./a.out '*.c'
I don't really know how to explain this but here's my problem:
I am trying to make my program accept command line arguments and then run a program via. the Linux command line (CentOS 6).
Heres the main function:
int main(int argc, char *argv[])
I am trying to run a Linux program, here's the code:
system("nmap -sT -p 19 1.1.1.* -oG - | grep 19/open > temp");
I want to replace '1.1.1.*' with the first argument I input into my C program, Ex:
system("nmap -sT -p 19 ", (argv[1]) "-oG - | grep 19/open > temp");
I have tried multiple ways and none seemed to work.
To sum it up, i'm trying to take the first argument I input into my program and use it in replace of the '1.1.1.*' in the system function. I have no idea on how to do this, I'm new to C programming. Thank you all replies are appreciated.
snprintf is the safest way to do this, this is a simple example without any checking of argc etc...:
#include <string.h>
#include <stdio.h>
int main(int argc, char *argv[])
{
char buf[200] ;
char str1[] = "nmap -sT -p 19 ";
char str2[] = " -oG - | grep 19/open > temp";
snprintf(buf, 200, "%s%s%s", str1, argv[1], str2);
printf( "%s\n", buf ) ;;
}
Just use the following syntax to accept command line arguments in Linux.
./program arg1 arg2