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Explain Operations involving unsigned and signed? unsigned will be converted to signed?
Answer:
yes,
void foo(void)
{
unsigned int a = 6;
int b = -20;
(a+b > 6) ? puts("> 6") :
puts("<=6");
}
Here output would give you "> 6". The reason for this is that expressions involving signed and unsigned types have all operands
promoted to unsigned types
void foo(void)
{
unsigned int a = 6;
int b = -20;
(a+b > 6) ? puts("> 6") :
puts("<=6");
}
Here output would give you "> 6". The reason for this is that expressions involving signed and unsigned types have all operands
promoted to unsigned types
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