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I am trying to create this little program to help me, with only one command, to compile and run a C program within Ubuntu's terminal.

Trying to make it fancier, I added an argument to the bash file so I can use it for any C program I want. So this is how it is supposed to go:

  • Create a variable to store the name of the file
  • Use that variable to compile the program (to the same file name)
  • Use that same name to run the file.

Here is the code:

# usr/bin/bash
filename=$1
cc -o $filename "$filename.c"
./$filename.out

almost everything runs, the only problem I still have is in the last line:

./$filename.out

It doesn't seem to use the name of the variable inside the command which executes the final program.

I'm a noob at bash (let's say I haven't used it in months).

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  • 4
    I guess the actual problem is that your cc command doesn't produce $filename.out but just $filename -- that being said, a shell script doesn't make a good build tool. You should read up on Makefiles instead. Commented Jul 26, 2017 at 11:14

3 Answers 3

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6

cc -o foo will output foo not foo.out. You should also double-quote the variable expansions to prevent IFS-splitting and globbing: 

filename=$1
cc -o "$filename" "$filename.c" &&
./"$filename"

Apart from that # /usr/bin/bash (unlike #!/usr/bin/bash) does nothing. It's a comment. The whole thing will be run the /bin/sh, not bash (but you don't need bash, anyway).

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5 Comments

I replaced # /usr/bin/bash by #!/usr/bin/bash and it didn't work as I expected. bash: ./runC_Script.sh: usr/bin/bash: bad interpreter: No such file or directory
@Chihab although it doesn't matter because you don't use any bash syntax -- you didn't put #!/usr/bin/bash but #!usr/bin/bash. You should stop being so sloppy.
I gotta look into this, seems important!
@Chihab You might not have bash in /usr/bin. It could be in /bin. I'd use /bin/sh (or /bin/sh -e to get aborts on failures by default) because it could be a much faster POSIX shell and a POSIX shell is all you need for this.
@Chihab as a general rule, prefer writing scripts with #!/bin/sh (they should therefore work with any sh-compatible shell and avoid bashisms). Only write #!/usr/bin/bash in scripts that explicitly need special bash syntax/features.
-1

You will need to use quotation marks:

./"$filename.out"
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1 Comment

This, although a good thing to do, is not necessary in every case. It definitely isn't the source of OP's problem.
-2 
#!/bin/sh
filename=$1
cc -o "${filename}.out" "${filename}.c"
"./${filename}.out"
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4 Comments

I could have used eval as well, still I needed to add .out in the first argument of cc as in cc -o "$filename.out" "$filename.c" Thanks :D
You need to use ${var} syntax to concatinate string
eval is completely unnecessary here .. the best you could achieve is that it doesn't actively hurt.
@FelixPalmen Yes, for single argument not necessary

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