What's wrong with this little program? I cannot get the correct answer. I just use m[1][1] to test but it's always 0!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int **m;
m = new int*[5];
for( int i = 0 ; i < 5; i++ )
{
m[i] = new int[5];
memset(m[i],1,sizeof(int)*5);
}
printf("%f",*(*(m+1)+1));
return 0;
}
This is "not C++". Yes, it's a C++ code, but it isn't using reasonable C++ idioms -- you're mixing C-style memset and pointer chasing with operator new (which is the only C++ feature you hapen to use). I will therefore assume that you have a reason to use code like that. If you don't have that, seriously consider using some STL class like a vector and avoid pointer manipulation.
If you used a reasonable compiler and added reasonable warnings, you would get the error immediately:
$ g++ -Wall pwn.cpp
pwn.cpp: In function ‘int main()’:
pwn.cpp:14:28: warning: format ‘%f’ expects argument of type ‘double’, but argument 2 has type ‘int’ [-Wformat]
The correct argument to printf for printing ints is %d. How does it look after changing that?
$ ./a.out
16843009
How come it doesn't print 1? The memset function sets memory, it does not initialize integers. And indeed, 0x01010101 == 16843009. This is not portable and very likely not what you want to do. Just don't do this, seriously.
for (int i = 0; i < 5; i++) is invalid in C, because you can't declare i there.
int i inside the for loop) is actually valid C99 code. You are right in that this was not allowed in earlier versions of the C standard, thanks for pointing that out. References: gcc says error: ‘for’ loop initial declarations are only allowed in C99 mode and note: use option -std=c99 or -std=gnu99 to compile your code.