Clarify how Rust treats backslashes at end of line in string literals
Rust differs in that behavior from C: In C, the newline escapes are resolved
before anything else, and in Rust this depends on whether the backslash is
escaped itself.
A difference can be observed in the following two programs:
```c
#include <stdio.h>
int main()
{
printf("\\
n\n");
return 0;
}
```
```rust
fn main() {
println!("\\
n");
}
```
The first program prints two newlines, the second one prints a backslash, a
newline, the latin character n and a final newline.
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@ -208,10 +208,10 @@ A _string literal_ is a sequence of any Unicode characters enclosed within two
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which must be _escaped_ by a preceding `U+005C` character (`\`).
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Line-break characters are allowed in string literals. Normally they represent
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themselves (i.e. no translation), but as a special exception, when a `U+005C`
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character (`\`) occurs immediately before the newline, the `U+005C` character,
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the newline, and all whitespace at the beginning of the next line are ignored.
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Thus `a` and `b` are equal:
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themselves (i.e. no translation), but as a special exception, when an unescaped
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`U+005C` character (`\`) occurs immediately before the newline (`U+000A`), the
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`U+005C` character, the newline, and all whitespace at the beginning of the
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next line are ignored. Thus `a` and `b` are equal:
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```rust
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let a = "foobar";
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