Why, warning: comparison between signed and unsigned integer expressions?

Hello all experts. If I compile the following code (small example) I
get "warning: comparison between signed and unsigned integer
expressions".

void test(void)
{
uint8_t a;
uint32_t i;

a = 10;

for(i = 0; i < (a + 1); i++)
{

}
}

But the following code doesn't give that warning:


void test(void)
{
uint8_t a;
uint32_t i;

a = 10;

for(i = 0; i < (uint32_t)(a + 1); i++)
{

}
}

What is going on here? Why does the compiler think (a + 1) is signed?
.

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