3

While using gets() in my code, the compiler shouts

warning: the 'gets' function is dangerous and should not be used.`

and

warning: ‘gets’ is deprecated (declared at /usr/include/stdio.h:638)
[-Wdeprecated-declarations]

Any specific reasons?

R Sahu
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network-programmer
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  • Use `fgets` with `stdin` instead. As the message says, `gets` is dangerous, since it provides no protection against buffer overflow.. – Weather Vane Jun 17 '15 at 11:49
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    The literal answer to "Why does the compiler emit a warning here?" is "Because the compiler's programmers made it do that." – user253751 Jun 17 '15 at 12:12

2 Answers2

6

Can someone explains why the compiler shows like that…?

Yes, because, the gets() function is dangerous, as it suffers from buffer overflow issue. Anyone should refrain from using that.

Also, regarding the warning with -Wdeprecated-declarations, gets() is no longer a part of C standard [C11 onwards]. So, C libraries compilers are not bound to support that any more. It can be removed in future. To warn the developer about the potential pitfall and to discourage the further usage of gets(), the compiler## emits the warning message.

(##) To be pedantic, the warning is not generated by the compiler (gcc) all by itself, rather caused by a pragma or attribute on the implementation of gets() in the glibc that causes the compiler to emit the warning. [Courtesy, FUZxxl, from the dupe answer.]

Sourav Ghosh
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2

  1. gets may cause buffer overflow, since it don't consider length of the data. More details are here : gets() function in C

  2. deprecated message means, this function is marked as deprecated and may remove from standard in later time. So discouraging user to use it.

Community
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Steephen
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