process.argv for command line arguments in Node.js - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Node.js script?

Consider this Node.js script run with the command: node script.js hello world. What will it print?

Node.js

console.log(process.argv.slice(2));

A ["hello world"]

B ["node", "script.js", "hello", "world"]

C ["script.js", "hello", "world"]

D ["hello", "world"]

💡 Hint

Remember that process.argv includes the node executable and script path as the first two items.

The process.argv array contains the node executable path at index 0 and the script path at index 1. Using slice(2) returns only the arguments passed after the script name, which are "hello" and "world".

❓ state_output

intermediate

What is the value of args after running this code?

If the script is run as node app.js --name=John --age=30, what is the value of args?

Node.js

const args = process.argv.slice(2);
console.log(args);

A ["--name=John", "--age=30"]

B ["name", "John", "age", "30"]

C ["--name", "John", "--age", "30"]

D ["app.js", "--name=John", "--age=30"]

💡 Hint

Arguments are passed as strings exactly as typed after the script name.

The process.argv array includes the arguments as strings exactly as typed. Since the arguments are passed as --name=John and --age=30, slicing from index 2 returns those exact strings.

📝 Syntax

advanced

Which option correctly extracts the third command line argument?

Given the command node index.js a b c d, which code correctly assigns the third argument to variable thirdArg?

A const thirdArg = process.argv[2];

B const thirdArg = process.argv[4];

C const thirdArg = process.argv.slice(2)[3];

D const thirdArg = process.argv.slice(3);

💡 Hint

Remember that process.argv[0] is node path, [1] is script path, so arguments start at index 2.

The process.argv array contains the node executable path at index 0 and the script path at index 1. Arguments start at index 2: 'a' (2), 'b' (3), 'c' (4), 'd' (5). The third argument is 'c' at index 4.

Option B gets the first argument ('a' at index 2).

Option B returns an array from index 3 onward (['b', 'c', 'd']).

Option B: process.argv[4] gets the third argument ('c').

Option B gets the fourth element of the sliced array starting at 2, which is 'd' (index 5 overall).

🔧 Debug

advanced

Why does this code print an empty array when run with arguments?

Given this code and running node app.js test, why does it print []?

Node.js

const args = process.argv.slice(3);
console.log(args);

A Because slice(3) returns the first three elements, which are empty.

B Because process.argv is empty when arguments are passed.

C Because slice(3) skips the first three elements, but only two exist after index 2.

D Because console.log cannot print arrays from process.argv.

💡 Hint

Check how many elements process.argv has when running with one argument.

process.argv has at least two elements: node path and script path. Arguments start at index 2. With one argument 'test', process.argv has length 3. Using slice(3) returns elements starting at index 3, which is beyond the last element, so it returns an empty array.

🧠 Conceptual

expert

What is the length of process.argv when running node without a script?

If you run node alone (no script, no arguments), what is the length of process.argv inside the Node.js REPL?

A 2

B 0

C 3

D 1

💡 Hint

Think about what process.argv contains by default: node path and script path.

Even when no script is provided, process.argv contains the path to the node executable at index 0 and the path to the REPL script at index 1, so its length is 2.