Can any body tell me ,why my last line prints the enteir string instead of an address
char str [20];
char *sp;
cin>>str;
sp=str;
cout<<&str[0];
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Kuttan Sujith
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`str` and `&str[0]` are equivalent from the `cout` point of view -- both are a pointer to the first char in `str`. – David C. Rankin Jul 12 '17 at 05:19
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Yes,but why it is spreading to entire string ? – Kuttan Sujith Jul 12 '17 at 05:21
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It is explained in the duplicate, you need to cast to a pointer `(void *)str` – David C. Rankin Jul 12 '17 at 05:22
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@KuttanSujith because the `operator<<(char*)` overload is designed to treat the pointer as a null terminated C string. You are looking for `operator<<(void*)` instead, which just prints the pointer address – Remy Lebeau Jul 12 '17 at 08:12