The correct answer is $-1$ . A quadratic equation in the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, has exactly one solution when its discriminant, $b^2 - 4ac$, is equal to $0$. In the given equation, $-16x^2 - 8x + c = 0$, $a = -16$ and $b = -8$. Substituting $-16$ for $a$ and $-8$ for $b$ in $b^2 - 4ac$ yields $(-8)^2 - 4(-16)(c)$, or $64 + 64c$. Since the given equation has exactly one solution, $64 + 64c = 0$. Subtracting $64$ from both sides of this equation yields $64c = -64$. Dividing both sides of this equation by 64 yields $c = -1$. Therefore, the value of $c$ is $-1$.