CODEWITHSUNDEEP
capibuffer-overflow
lsalamon
lsalamon

Reputation: 8164

This form of use of printf api makes it safer?

char str[] = "some text";

printf ( "%.*s", strlen(str), str );

** Of course, their buffers, strings yet to be properly targeted

Upvotes: 1

Views: 357

Answers (4)

nobody
nobody

Reputation: 20163

No, it's no different than:

char str[] = "some text";
printf("%s", str);

Upvotes: 0

Dathan
Dathan

Reputation: 7446

In the example you give, there's no difference. printf interprets the undecorated "%s" character code as meaning "read and print all characters from a character pointer until a null character is encountered." The initializer char str[] = "some text"; automatically appends the null character, so there will be no overrun. On the other hand, the following is not safe

char str[] = {'h', 'e', 'l', 'l', 'o', 'w', 'o', 'r', 'l', 'd'};
printf("%s", str);

because no null character is appended to the character sequence str. On the other hand, since strlen(str) determines string length by counting the number of characters before a null is encountered, it doesn't offer you any benefit over just using printf without a field width specified.

The upshot: the only case where specifying a field width for a string is helpful is when the string isn't guaranteed to be null-terminated (or the classic case of using sprintf to write to a buffer that may not be big enough to hold the contents of str), but in that case you'd have to determine string length using something other than strlen.

Upvotes: 1

Roger Pate
Roger Pate

Reputation:

No, that just shifts the problem of detecting the end of the string from printf to strlen, and it's still exactly the same.

Upvotes: 3

Martin Beckett
Martin Beckett

Reputation: 96119

No, printf should be safe form overruns anyway setting a field width doesn't really help

Upvotes: 1

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