I tried to execute script within C program. I tried:
system("/home/olaudix/weather.sh")
system("sh /home/olaudix/weather.sh")
execvp("/home/olaudix/weather.sh")
but all of them throw
syntax error: "(" unexpected at line 1.
Scripts starts with function getData() { but it runs fine when executed in terminal.
The shebang (i.e.: #!) at the very beginning of the scripts is missing. You need it in order to specify the corresponding interpreter (bash in this case), i.e.:
#!/bin/bash
It is also possible to source the shell script in the system command. Below is an example
$ cat 48465591.c
#include<stdio.h>
#include<stdlib.h>
int main(int argc,char* argv[])
{
system(". ./myshellscript.sh"); // Note the '.' which stands for sourcing
return 0;
}
$ cat myshellscript.sh
printhello()
{
echo "Hello";
}
Related
The code is given below.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main ( int argc, char *argv[] )
{
//FILE *fps;
char secret[512] =" ";
FILE *fps = fopen("/etc/comp2700/share/secret", "r");
if(fps == NULL)
{
printf("Secret file not found\n");
return 1;
}
fgets(secret, 512, fps);
printf("Secret: %s\n", secret);
fclose(fps);
return 0;
}
When I am trying to run this program it is repeatedly throwing the following error:
./attack1.c: line 4: syntax error near unexpected token `('
./attack1.c: line 4: `int main ( int argc, char *argv[] )'
You need to compile your source file with gcc as follows
gcc -o attack attack1.c
then run it with
./attack
You should read up on the difference between compiled versus interpreted languages.
There is a short video here explaining the difference.
You cannot run your C program from the command line as ./attack1.c. Normally the shell would refuse to execute the C source file because it should not have execute permission, but for some reason, on your system, it must have the x bits and is read by the default shell as a script.
Of course this fails because attack1.c contains C code, not a command file. Note that the #include lines are interpreted as comments by the shell and the error only occurs at line 4.
To run a C program, you must first compile it to produce an executable:
gcc -Wall -o attack1 attack1.c
And then run the executable if there were no compilation errors:
./attack1
You can combine these commands as
gcc -Wall -o attack1 attack1.c && ./attack1
First, you need to compile the attack.c code using the following command:
gcc attack.c
This will create one executable file a.out which you can run using the following command:
./a.out
Hope this helps you.
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.
I'm fairly new to C and am completely new to using the command prompt and GCC to compile and run my programs. I'm struggling to find the right words to ask this question properly so please bear with me, I am doing my best.
I need to use GCC to compile and run this C program but I'm getting an error that I do not understand. In this example program, I was told to use these lines to compile and run the code:
$ gcc -Wall -std=c99 -o anagrams anagrams.c
$ ./anagrams dictionary1.txt output1.txt
So that is what I did. GCC does compile the program file, so the first line does not give me any error. But GCC does not like the second line, as shown below:
C:\Users\...\Example>gcc -Wall -std=c99 -o anagrams anagrams.c
C:\Users\...\Example>./anagrams dictionary1.txt output1.txt
'.' is not recognized as an internal or external command,
operable program or batch file.
Everywhere I look, it says to use "./filename" to run the program after compiling so I don't understand why it is not working for me. Any help or advice would be really appreciated.
Also, here is the main() of the program to show why those two .txt files are needed:
int main(int argc, char *argv[])
{
AryElement *ary;
int aryLen;
if (argc != 3) {
printf("Wrong number of arguments to program.\n");
printf("Usage: ./anagrams infile outfile\n");
exit(EXIT_FAILURE);
}
char *inFile = argv[1];
char *outFile = argv[2];
ary = buildAnagramArray(inFile,&aryLen);
printAnagramArray(outFile,ary,aryLen);
freeAnagramArray(ary,aryLen);
return EXIT_SUCCESS;
}
'.' is not recognized as an internal or external command
This is not a GCC error. The error is issued by the shell when trying to run a command.
On Windows this
./anagrams dictionary1.txt output1.txt
should be
.\anagrams dictionary1.txt output1.txt
as on Windows the path delimiter is \ as opposedto IX'ish systems where it is /.
On both systems . denotes the current directory.
The reason for the crash you mention in your comment is not obvious from the minimal sources you show. Also this is a different question.
I have shell scripts and I need to run that continuous work in background.
For example:
#include <stdio.h>
int main(int argc, char **argv)
{
for (; ;) {
system("./dup -r /root/duptest/");
sleep(60);
}
return 0;
}
It's working and run every minute.
First question: How can I run this background(like & --> ./dup ... &) without put &.
Second question: How can I put shell codes in C source codes?
I found this, Do I need to put \n\ for all lines? It's so hard for edit.
#include <stdio.h>
#include <stdlib.h>
#define SHELLSCRIPT "\
#/bin/bash \n\
echo \"hello\" \n\
echo \"how are you\" \n\
echo \"today\" \n\
"
int main()
{
system(SHELLSCRIPT);
return 0;
}
Third question: How can I use shell parameter in C, like this:
./dup.exe -r /blablabla...
mean
system("./dup -r /blablabla");
I need to use $1 $2 parameter with compiled C program.
Question 1: Look for "how to make a process as deamon process in UNIX" Although daemon process is a overkill for your requirement, you can perform steps until the process is running according to your requirements
Question 3: You need to have command line arguments, check about that. Your main should look like main(int arg_count, char *args_vector[]){...} and in that you can access each command line argument as an array element
Q1: use fork() and don't wait on the child's PID.
Q2: C and C++ will concatenate adjacent string literals, like so:
static const char script[] =
"echo hello\n"
"echo how are you\n"
"echo today"
;
int main(int argc, char* argv[])
{
puts(script); // so you can see what it looks like
// system(script); // <-- uncomment this line to actually run it.
return 0;
}
Q3: use the argc and argv parameters to main() to build the command line you want to execute.
I have two (Ubuntu Linux) bash scripts which take input arguments. They need to be run simultaneously. I tried execve with arguments e.g.
char *argv[10] = { "/mnt/hgfs/F/working/script.sh", "file1", "file2", NULL };
execve(argv[0], argv, NULL)
but the bash script can't seem to find any arguments at e.g. $0, $1, $2.
printf "gcc -c ./%s.c -o ./%s.o\n" $1 $1;
gcc -c ./$1.c -o ./$1.o -g
exit 0;
output is gcc -c ./main.c -o ./main.o
and then a lot of errors like /usr/include/libio.h:53:21: error: stdarg.h: No such file or directory
What's missing?
Does your script start with the hashbang line? I think that's a must, something like:
#!/bin/bash
For example, see the following C program:
#include <stdio.h>
#include <unistd.h>
char *argv[10] = { "./qq.sh", "file1", NULL };
int main (void) {
int rc = execve (argv[0], argv, NULL);
printf ("rc = %d\n", rc);
return 0;
}
When this is compiled and run with the following qq.sh file, it outputs rc = -1:
echo $1
when you change the file to:
#!/bin/bash
echo $1
it outputs:
file1
as expected.
The other thing you need to watch out for is with using these VMWare shared folders, evidenced by /mnt/hgfs. If the file was created with a Windows-type editor, it may have the "DOS" line endings of carriage-return/line-feed - that may well be causing problems with the execution of the scripts.
You can check for this by running:
od -xcb /mnt/hgfs/F/working/script.sh
and seeing if any \r characters appear.
For example, if I use the shell script with the hashbang line in it (but appen a carriage return to the line), I also get the rc = -1 output, meaning it couldn't find the shell.
And, now, based on your edits, your script has no trouble interpreting the arguments at all. The fact that it outputs:
gcc -c ./main.c -o ./main.o
is proof positive of this since it's seeing $1 as main.
The problem you actually have is that the compiler is working but it cannot find strdarg.h included from your libio.h file - this has nothing to do with whether bash can see those arguments.
My suggestion is to try and compile it manually with that command and see if you get the same errors. If so, it's a problem with what you're trying to compile rather than a bash or exec issue.
If it does compile okay, it may be because of the destruction of the environment variables in your execve call.