Question

 u32 iterations = 5;
 u32* ecx = (u32*)malloc(sizeof(u32) * iterations);

 memset(ecx, 0xBAADF00D, sizeof(u32) * iterations);
 printf("%.8X\n", ecx[0]);

 ecx[0] = 0xBAADF00D;
 printf("%.8X\n", ecx[0]);

 free(ecx);

Very simply put, why is my output the following?

0D0D0D0D
BAADF00D

ps:
u32 is a simple typedef of unsigned int

edit:

• Compiling with gcc 4.3.4
• string.h is included

Answer 1

The second parameter to memset is typed as an int but is really an unsigned char. 0xBAADF00D converted to an unsigned char (least significant byte) is 0x0D, so memset fills memory with 0x0D.