Is photosynthesis endothermic and exothermic at the same time? ... somehow? Photosynthesis is obviously an endothermic reaction, I mean, what would be the point otherwise, right? This is probably just sheer and utter stupidity on my behalf, but why does Wikipedia say the heat of formation of glucose is negative? As you can probably see I'm on a quest of understanding enthalpy as a concept and I'm still quite new, but some sort of ultranoob clarification would be very much welcome.
Consider that when $1$ molecule of glucose forms, this happens at the cost of $6$ molecules of  $\ce{CO2}$ and $6$ molecules of $\ce{H2O}$ :

$
\ce{6CO2(g) + 6H2O(l) -&gt; C6H12O6(s) + 6O2(g)}
$

Considering that at $25\ \mathrm{^\circ C}$ and  $1\ \text{atm}$

$
\Delta H_{\text{f},\ce{CO2}}^\circ=-393.5\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{H2O}}^\circ=-285.8\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{C6H12O6}}^\circ=-1273.3\ \text{kJ/mol}
$

You get, by summing the enthalpies of formation for each of the involved species and considering that for $\ce{O2}$ that enthalpy is zero:

$
\Delta H_\text{r}^\circ=6\times393.5\ \text{kJ/mol}+6\times285.8\ \text{kJ/mol}-1273.3\ \text{kJ/mol}=2802.5\ \text{kJ/mol}
$

Note that, for $\ce{CO2}$ and $\ce{H2O}$, the change of sign is required, since you "lose" them, while you form glucose.

So, finally $\Delta H_\text{r}^\circ$ (heat to be absorbed to form one mole of glucose) is positive, indicating that photosynthesis is indeed endothermic.