Converting char* string to lowercase using tolower() causes SIGSEGV - what's wrong?

Question

I've found the following code in a legacy project that seems to cause some trouble:

for( ; *str; ++str )
    *str = tolower(*str);

The trouble is, we get an SIGSEGV at the moment the result of tolower() should be written back to *str . This is what I got from using printf() debugging and the stracktrace produces, as we're using JNI here and thus attaching a debugger to the C libraries is not working for us.

The code fails if it's compiled with gcc on linux. Running the same thing on windows (CMake is being used for this cross-plattform setup) works fine.

Where can I look to find the reason for this? Any hints appreciated :)

thank you - Markus

What Michael wanted to say is, modifying string constants can lead to segmentation faults. Here is why: http://stackoverflow.com/questions/164194/why-do-i-get-a-segmentation-fault-when-writing-to-a-string — ArjunShankar, Apr 10 '12 at 09:30
Yes, that's what I (and later @ouah in the anwer) wanted to say ;-) — Michael Krelin - hacker, Apr 10 '12 at 09:31
Hm, not really :( But maybe I got it now: the caller might be passing a pointer into myfunction that - for some reason - cannot be written to. Can I check this? I'll try to trace down where the given string comes from. Thanks a lot for your fast replies! — Markus, Apr 10 '12 at 09:36

score 2 · Accepted Answer · answered Apr 10 '12 at 09:29

2

String literals are non-modifiable:

char blop[] = "modifiable string";
char *bla = "non-modifiable string";

*blop = tolower(*blop); // OK
*bla = tolower(*bla);   // not OK, modifying a string literal

answered Apr 10 '12 at 09:29

ouah

142,963
15
272
331

+1, but `tolower()` function doesn't expect `char*` and doesn't modify anything ;-) What you meant is what the OP refers to as `myfunction()` in comments above. – Michael Krelin - hacker Apr 10 '12 at 09:30

Converting char* string to lowercase using tolower() causes SIGSEGV - what's wrong?

1 Answers1