When I declared a variable
char buf[512];
What does mean &buf ?
It seems equals to buf :
printf(" buf : %p %p\n", (void *) buf, (void *) &buf);
prints :
buf : 0x7ffda6053fe0 0x7ffda6053fe0
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Sander De Dycker
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Robert Po
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4What does `char[512] buf` mean? Is it a typo error? – Jose Sep 12 '18 at 07:39
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Not only is that not valid C, if you want to print out a pointer's value in `printf()` and friends, `%d` is not the right format specifier... go back and read the documentation on the function. And see if you can find something about pointer decay while you're doing so. – Shawn Sep 12 '18 at 07:44
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Do you mean char buf[512]; ? – Mike Sep 12 '18 at 07:44
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(http://c-faq.com/aryptr/index.html might be useful reading) – Shawn Sep 12 '18 at 07:45
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Yes,I mean char buf[512]. – Robert Po Sep 12 '18 at 07:46
2 Answers
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it is the address of the 1st element in the array (i.e. address of buf[0]) vs. the address of the array itself (i.e. buf). No big surprise that they are equal.
CIsForCookies
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The address is the same because an array decays into a pointer to the first element when passed, on the other hand, the address-of operator & takes the address of the array, the address of the first element and the address of the array itself are the same.
Use %p instead of %d to print addresses:
printf(" buf : %p %p\n", (void *)buf, (void *)&buf);
David Ranieri
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