带空格的C程序命令行参数

Question

My python program program.py prints text with spaces, and my C program a.out is only expecting a single command line argument. Why is there different behavior between these two methods... ie why does Method 1 work and not Method 2?

Method 1:

# program.py
print("Hello world")

Terminal:

> ./a.out "$(python program.py)"
// Program successfully run

Method 2:

# program.py
print("\"Hello world \"")

Terminal:

> ./a.out $(python program.py)
// Error- only one command line argument expected

Answer 1

Method 1:

./a.out "$(python program.py)"

./a.out will use "$(python program.py)" as a string
echo "$(python program.py)" prints "$(python program.py)"

Method 2:

./a.out $(python program.py)

./a.out will use the result of $(python program.py) as an argument.
echo $(python program.py) will run the program, and what is printed will be used as an argument.

Because you have printed Hello world , it will use Hello as the first argument, and world as the second argument. it will result in something like
./a.out Hello World

Answer 2

Apparently whenever you use a $something, it will always be split in several parameters if you have spaces in it. Just look at how $1 works, it's the same behavior. Then your $(command) will work the same. This is why you have to put quotes around it.

I tried:

ls $(echo '"Hello World!"')

It has the same behavior than your second method.

And this:

ls "$(echo 'Hello World!')"

Is the same than your first method.