[Solved] Usage of uint_8, uint_16 and uint_32

In your 3 cases, before the printf statement, the 4 first bytes of the arr array (in hexadecimal format) are: FF D8 FF E0, which corresponds to 255 216 255 224. Here are explanations of each case: arr[0] has uint8_t type, so its 1-byte value is 0xFF, so printf(“%i”, arr[0]); prints 255, which corresponds to … Read more