CODEWITHSUNDEEP
shell
maulzey
maulzey

Reputation: 4240

Where am I screwing up

I want to time in seconds for execution of shell script.

My implementation:

#!/bin/sh
START=$(date +%s)
echo $START
.
.
bla bla bla
.
.
.
END=$(date +%s)
echo $END
DIFF=($END - $START)
echo  "Time difference is "$DIFF

This is showing me END time and not difference. I cannot guess the reason why minus is not working.

Upvotes: 1

Views: 73

Answers (2)

Jonathan Leffler
Jonathan Leffler

Reputation: 753725

You need dollar double parentheses for arithmetic:

DIFF=$(($END - $START))

As in:

#!/bin/bash
START=$(date +%s)
echo $START
sleep 2
END=$(date +%s)
echo $END
DIFF=$(($END - $START))
echo  "Time difference is" $DIFF

Bash Reference Manual says:

3.5.5 Arithmetic Expansion

Arithmetic expansion allows the evaluation of an arithmetic expression and the substitution of the result. The format for arithmetic expansion is:

$(( expression ))

The expression is treated as if it were within double quotes, but a double quote inside the parentheses is not treated specially. All tokens in the expression undergo parameter expansion, command substitution, and quote removal. Arithmetic expansions may be nested.

See also:

3.2.4.2 Conditional Constructs

((…))

(( expression ))

The arithmetic expression is evaluated according to the rules described below (see Shell Arithmetic). If the value of the expression is non-zero, the return status is 0; otherwise the return status is 1. This is exactly equivalent to

let "expression"

See Bash Builtins, for a full description of the let builtin.

Upvotes: 1

Tim Pote
Tim Pote

Reputation: 28029

Assuming you're using bash:

You need to use double parens to do arithmetic in the shell:

((DIFF = START - END))
#or
DIFF=$((START - END))

Note that you don't have to use dollar signs to read from variables inside of double parens.

Upvotes: 2

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