C Programming - Size of 2U and 1024U

Question

I know that the U literal means in c, that the value is a unsigned integer. An unsigned intagers size is 4 bytes. But how big are 2U or 1024U? Does this simply mean 2 * 4 bytes = 8 bytes for example or does this notation means that 2 (or 1024) are unsigned integers? My goal would be to figured out how much memory will be allocated if i call malloc like this int *allocated_mem = malloc(2U * 1024U); and prove in a short program my answer what i tried like this printf("Size of 2U: %ld\n", sizeof(2U)); printf("Size of 1024U: %ld\n", sizeof(1024U)); I would have expeted for the first line a size of 2 * 4 Bytes = 8 and for the second like 1024 * 4 Bytes = 4096 but the output is always "4". Would realy appreciate what 2U and 1024U means exactly and how can i check their size in C? JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

My goal would be to figured out how much memory will be allocated if i call malloc like this int *allocated_mem = malloc(2U * 1024U); What is difficult about 2 * 1024 == 2048? The fact that they are unsigned literals does not change their value. An unsigned intagers size is 4 bytes. (sic) You are correct. So 2U takes up 4-bytes, and 1024U takes up 4-bytes, because they are both unsigned integers. I would have expeted for the first line a size of 2 * 4 Bytes = 8 and for the second like 1024 * 4 Bytes = 4096 but the output is always "4". Why would the value change the size? The size depends only on the type. 2U is of type unsigned int, so it takes up 4-bytes; same as 50U, same as 1024U. They all take 4-bytes. You are trying to multiply the value (2) times the size of the type. That makes no sense.