I created the following array:
char* myarr= "data successfully inserted";
when I do
sizeof(myarr)
I get 8. Why do I get that?. If 1 char is 1 byte then size of myarr should be 26 by just counting the characters right?
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John Kugelman
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daniel
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1Because you are getting the size of the *pointer.* You should have declared it as `char[] myarray = "data successfully inserted";`, which would save 8 bytes of space. – user207421 Apr 16 '17 at 06:50
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1If the expression `e` has the type `T`, `sizeof(e)` is equivalent to `sizeof(T)`. All objects of a given type have the same size. – molbdnilo Apr 16 '17 at 06:51
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2Because you are abusing `sizeof`, and should be using `std::string` first, followed by raw character arrays only if you really need them. – StoryTeller - Unslander Monica Apr 16 '17 at 06:53
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just do `std::string mystr = "..."` and use `mystr.size()` instead – Ap31 Apr 16 '17 at 07:00
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2"All objects of a given type have the same size". You should write this on a blackboard 100 times. – n. 1.8e9-where's-my-share m. Apr 16 '17 at 07:09
3 Answers
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On your platform, a char pointer is 8 bytes. That's the size of the pointer variable, not the size of whatever it's pointing to. If you want the length of a string, call strlen(myarr).
John Kugelman
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Because that's the size of a pointer on your system. If you want to get the length of a string, you should use strlen() function.
msc
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sizeof(myarr) is not the size of the string but the size of char * pointer on your platform which gives you 8 bytes. the size of the array can be calculated by strlen(myarr)
Shubham Khatri
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