1 SC2307
Vidar Holen edited this page 2021-08-22 20:28:54 -07:00

'expr' expects 3+ arguments but sees 1. Make sure each operator/operand is a separate argument, and escape <>&|.

Problematic code:

# | not escaped
expr 1 | 2
# > not escaped
expr "$foo" >= "$bar"

# Missing spaces around +
expr 1+2
# Unexpected quoting around an expression
expr "1 + 2"

Correct code:

expr 16 \| 7
expr "$foo" \>= "$bar"
expr 1 + 2

Rationale:

ShellCheck found an expr command with 1 or 2 arguments. expr normally expects 3 or more.

Generally, this happens for one of two reasons:

  • You are using an operator like |, &, >, >=, <, <=, which needs to be escaped to avoid the shell interpreting it as a pipe, backgrounded command, or redirection.
  • You don't have spaces around operators and operands (or have bad quotes) which causes them not to be separate arguments.

Make sure each operator or operand to expr is a separate argument, and that anything containing shell metacharacters is escaped. The correct code shows examples of each.

Exceptions:

None

  • Help by adding links to BashFAQ, StackOverflow, man pages, POSIX, etc!