I need to pass an argument which will change every time from C program to a shell script.
int val=1234;
char buf[100];
sprintf(buf,"echo %d",val);
system("call.sh $buf");
call.sh::
#!/bin/sh
echo "welcome"
echo $*
echo "done"
output of C is::
welcome
done
I cant see the argument value which is 1234 in the script. Can anybody suggest me to get right value...
You can't pass a C variable as a shell variable. You need to build the whole command line in the string, and then pass it to system(...)
int val=1234;
char buf[100];
sprintf(buf, "call.sh %d", val);
system(buf);
You should use the setenv(), getenv() or putenv() functions (defined instdlib.h). Quoting the man:
The
setenv()function adds the variable name to the environment with the value value, if name does not already exist. If name does exist in the environment, then its value is changed to value if overwrite is nonzero; if overwrite is zero, then the value of name is not changed. This function makes copies of the strings pointed to by name and value (by contrast withputenv(3)).
The prototype of the function is the following:
int setenv(const char *name, const char *value, int overwrite);
