I must first log a variable using console.log, then execute a function on the same line, followed by logging the variable again

Essentially, I have a variable called c1 that is assigned a random hexadecimal value. After printing this to the console, I want to print another hex value without creating a new variable (because I'm feeling lazy). Instead, I intend to achieve this through the following method:

function thing() {
  // code to generate a random hex value (which currently works)
}

console.log(variable + thing() + variable);

I have yet to find a solution for this problem and I am unsure if it's even feasible. If not, I will resort to using additional variables, but I prefer to give this method a shot first.

Feeling a bit lost on where to start, so here goes nothing!

Answer №1

This query is quite unconventional, and the exact intention behind it is unclear. Allow me to provide a suggestion to streamline the process. To combine invoking a function and calling console.log in a single line while also including a function within the console log statement, consider using multi-line literals with templates as shown below:

let value = something()
console.log( `value = ${value}
something() = ${something()}
value = ${value}`)

By utilizing this approach, each variable and function call will be displayed on a separate line in the console, giving the appearance of distinct console.log calls despite being a single call fundamentally.

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