I have this code that gives a pointer an address and print it, but why that does not work??
void main()
{
int *b = (int*) 32;
printf("%d\n",b[0]);
}
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4How does it not work? – Zoltán Jul 14 '13 at 10:59
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2I ran the code and it formatted my hard drive. – Thomas Jul 14 '13 at 11:05
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I am voting to close - I don't know what "does not work" mean. – Aniket Inge Jul 14 '13 at 11:06
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You are assigning fake address read: [Invalid address causes Undefined behavior at runtime](http://stackoverflow.com/questions/17627394/why-do-we-store-string-in-a-character-pointer-e-g-in-fopen/17627415#17627415) – Grijesh Chauhan Jul 14 '13 at 11:10
2 Answers
4
b[0] dereferences an array which points to memory you haven't allocated. The effects of doing this are undefined. You might get a value returned or your program may crash if address 32 isn't readable from your process.
simonc
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Its platform dependant. If the OS decides the address is not readable, there's nothing you can do. Why do you want to read this particular location in memory? Or do you want to print the address of b rather than what it points to – simonc Jul 14 '13 at 11:28
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int *b = (int*) 32;
above code assigns memory address 32 to this pointer, i don't think that is you want, you will get access denied error when you call printf, hope the following codes is useful toyou
int a = 32;
int *b = &a;
printf("%d\n",b[0]);
//output 32
printf( "%d\n", &b);
// output b pointer address.
Jenny 1985
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1In addition to the last one being %p, I expect the intention is to print `b` rather than `&b`. – mah Jul 14 '13 at 11:22
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if you wanna change the address, int *b = (int*) 32; is correctly， but (int*)32 is invalid pointer address, you should assign a valid pointer address to it. – Jenny 1985 Jul 14 '13 at 12:47