Char 'apex' C++ -

i building simple program use dynamic array in c++, program read file , after find word , when find change lower case upper case. find word control if after , before there 1 space or punctuation char.

when control if there ' (apex) have problem:

s[i+j-1] == '''

this because second close first, , third open char.

it run if use ascii code: s[i+j-1] == 39.

how can write program without using ascii code?

' needs escaped in character literal:

s[i+j-1] == '\''