On 27/11/2019 08:47, Raphael “Teemperor” Isemann via lldb-dev wrote:
This can also be reproduced in the command line like this:
(lldb) expr "foo"
(const char [4]) $0 = "foo"
(lldb) expr "bar"
(const char [4]) $1 = "bar"
(lldb) expr $0
(const char [4]) $0 = “bar”
This however works just fine:
(
This can also be reproduced in the command line like this:
(lldb) expr "foo"
(const char [4]) $0 = "foo"
(lldb) expr "bar"
(const char [4]) $1 = "bar"
(lldb) expr $0
(const char [4]) $0 = “bar”
This however works just fine:
(lldb) expr char c[] = "foo"; c
(char [4]) $0 = "foo"
(lldb) expr char c
