signed

I still haven't found a reason why the lowest signed negative number doesn't have an equivalent signed positive number? I mean in a 3 digit binary number for simplicity 100 is -4? but we can't have a positive 4 in signed format because we can't. It overflows. So how do we know two's complement 1000 is -4 1000 0000 is -128 and so on? We have no original positive number One way to think about it is that signed, two's complement format works by assigning each bit a power of two, then flipping the sign of the last power of two. Let's look at -4, for example, which is represented as 100. This means

This question already has an answer here: Comparison operation on unsigned and signed integers 7 answers int main(void) { unsigned int y = 10; int x = – 4; if (x > y) Printf("x is greater"); else Printf("y is greater"); getch(); return (0); } Output: x is greater I thought the output would be y is greater since it is unsigned. What's the reason behind this? Because the int value is promoted to an unsigned int . specifically 0xFFFFFFFC on a 32-bit machine, which as an unsigned int is 4294967292 , considerably larger than 10 C99 6.3.1.1-p2 If an int can represent all values of the original type

I have below a simple program: #include <stdio.h> #define INT32_MIN (-0x80000000) int main(void) { long long bal = 0; if(bal < INT32_MIN ) { printf("Failed!!!"); } else { printf("Success!!!"); } return 0; } The condition if(bal < INT32_MIN ) is always true. How is it possible? It works fine if I change the macro to: #define INT32_MIN (-2147483648L) Can anyone point out the issue? Lundin This is quite subtle. Every integer literal in your program has a type. Which type it has is regulated by a table in 6.4.4.1: Suffix Decimal Constant Octal or Hexadecimal Constant none int int long int unsigned