bash scripting problem
 

hey there
i was trying to execute a c program using a bash shell script
script was as follows

#!/bin/bash
`gcc shiv.c -o shiv`
`chmod 777 shiv`
`./shiv`


this was displaying an error that enter command not found.(ENTER is written in the first printf command in shiv.c file.


then i tried
#!/bin/bash
echo `gcc shiv.c -o shiv`
echo `chmod 777 shiv`
echo `./shiv`
then the problem was that i had to enter all the inputs first then the output was coming.
e.g in my c program

main()
{
int a;
printf("enter the no of persons");
scanf("%d",&a);
}
when i executed the script written three commands as mentioned above the scanf statement was running before printf;


kindly help me..

The main problem here is the weird usage of backticks, that mean command substitution: the ./shiv command is executed and the output is executed again, hence the error you get. Why did you use backticks?
Moreover the chmod statement is not needed, since gcc produces an executable with the proper permissions.

You don't need the back-ticks, try this:EDIT: "Beaten" by colucix......

just to make you understand backticks are used for a purpose generally to store an output from a command not anywhere.

echo is itself a "command" and then you are using backticks under it for chmod too, that is never required.

give some time to read google pages and get the correct use of commands before actually using them in scripts.