variables - Why compiler is not giving error when signed value is assigned to unsigned integer? - C++

Question

I know unsigned int can't hold negative values. But the following code compiles without any errors/warnings.

unsigned int a = -10;

When I print the variable a, I get a wrong value printed. If unsigned variables can't hold signed values, why do compilers allow them to compile without giving any error/warning?

Any thoughts?

Edit

Compiler : VC++ compiler

Solution

Need to use the warning level 4.

Answer 1

Microsoft Visual C++:

warning C4245: 'initializing' : conversion from 'int' to 'unsigned int', signed/unsigned mismatch

On warning level 4.

G++

Gives me the warning:

warning: converting of negative value -0x00000000a' to unsigned int'

Without any -W directives.

GCC

You must use:

gcc main.c -Wconversion

Which will give the warning:

warning: negative integer implicitly converted to unsigned type

Note that -Wall will not enable this warning.

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Maybe you just need to turn your warning levels up.