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lang-c
printf("%c", 1["Hello"]). You might be surprised. The swap1and the string literal and try again. Note: That is just to make you think - do not use this in your code. You will find the answer here already, so please do not add a new one.char s[] = "my string";andsis right-sized to 10. 2)&sis the address of a pointer to an array of 10char- not the same type aschar *.sis a variable, as such it needs memory (as any normalintwould). It happens that it points to an address, but thats just a detail. No additional space is required, if you lose that address, its lost for good, no one keeps track of it. Then,sis a variable (that consumes 4 bytes), that points to a piece of memory that has 10 bytes. Hope this clarifies a bit.