Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
I recently installed a 64bit OS on my computer, I thought that sizeof(char*) would give me 8 instead of 4. Shouldn’t I get a 64 bit addresses in my pointers?
This is because the compiler you are using is emitting 32 bit code. If you use a 64 bit compiler then pointers will be 8 bytes wide.
Note that most 64 bit systems have the ability to run 32 bit code under an emulation layer. On Windows the emulation layer is known as WOW64. This is clearly what is happening here.