i have not created the array name then why this C program is working.`
#include<stdio.h>
int main()
{
printf("%c",7["sundaram"]);
return 0;
}
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1Because `a[b]` is the same as `b[a]`. There are duplicates out there. – juanchopanza Oct 09 '14 at 07:04
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@juanchopanza but "sundaram" is not array name.it is just a content in that array.and i have not mention array name also – Walter Oct 09 '14 at 07:05
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1But you can apply `operator[]` to a pointer, and "sundaram" decays to pointer to `char`. – juanchopanza Oct 09 '14 at 07:06
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2@Walter: the string literal decays into something like a `const char*`. – Michael Oct 09 '14 at 07:06
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More like a `const char[8]` – dom0 Oct 09 '14 at 08:28