135. str::strip_prefix — Trim a Prefix Without Slicing by Hand

Reaching for if s.starts_with("foo") { &s[3..] } to drop a prefix? That’s an off-by-one waiting to happen — and a panic the first time someone passes in an emoji. str::strip_prefix returns Option<&str> and gets it right by construction.

The Problem

You want the part of a string after a known prefix:

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let s = "Bearer abc123";

let token = if s.starts_with("Bearer ") {
    &s[7..]
} else {
    s
};
assert_eq!(token, "abc123");

Two things wrong here: the literal 7 has to stay in sync with the literal "Bearer ", and slicing by byte offset will panic if the prefix ever lands mid-codepoint. Even using prefix.len() only saves you from the first bug, not the second when the prefix is dynamic.

The Fix: strip_prefix

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let s = "Bearer abc123";

let token = s.strip_prefix("Bearer ").unwrap_or(s);
assert_eq!(token, "abc123");

strip_prefix returns Some(&str) if the prefix matched (giving you the rest), or None if it didn’t. No magic numbers, no slicing, no UTF-8 footguns — the prefix length comes from the prefix itself.

Pattern Matching, Not Just Strings

The argument is anything implementing Pattern, so a char, a closure, or even an array of chars all work:

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assert_eq!("-x".strip_prefix('-'), Some("x"));
assert_eq!("x".strip_prefix('-'), None);

// Trim any leading whitespace character
let s = "\t  hello".strip_prefix(|c: char| c.is_whitespace());
assert_eq!(s, Some("  hello"));

Note this only strips one match — the char form doesn’t loop. For “strip every leading space,” reach for trim_start_matches.

The Twin: strip_suffix

Same shape, other end:

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let filename = "report.tar.gz";

let stem = filename.strip_suffix(".gz").unwrap_or(filename);
assert_eq!(stem, "report.tar");

Together they replace half the manual &s[..s.len() - 3] arithmetic you’d otherwise write — and the Option return makes “did it actually have the prefix?” a value, not a separate starts_with call.