Why And the value of uint32 with 0xFFFFFFFF?

Question

My friend was looking at this open source SSL code in the SHA encryption processing functions and noticed this odd piece:

ctx->total[0] += (uint32_t) ilen; // ilen is of type size_t ctx->total[0] &= 0xFFFFFFFF; if( ctx->total[0] < (uint32_t) ilen ) ctx->total[1]++; 

We cannot understand two things about this code. Firstly, it is ANDs ctx->total[0] (like uint32_t ) with 0xFFFFFFFF , which should do nothing. In binary format, this ANDing with all 1 s, which should give the same value. In my opinion, these two lines are identical:

 ctx->total[0] &= 0xFFFFFFFF; ctx->total[0] = ctx->total[0]; 

If I am right, why this line? Some reason for security? If I am wrong, how and why?

Secondly, we do not understand when this if ever will be true, if we assume that And does nothing. If And does nothing, then if essentially:

 if (ctx->total[0] < ctx->total[0]) 

which should never be true. What are we missing?

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If you want to see the header file to make sure ctx->total[0] is of type uint32_t or for some other reason, you can find it here.

Also, my first wild guess is that something is hiding there when we drop ilen from size_t to uint32_t , but I'm still stuck and confused.

Answer 1

First question:

You are right that this is not necessary for 32-bit, I think they are trying to prevent situations where ctx->total[0] not 32 bits (so even if someone changes it or the platform will have 64-bit for type uint32_t), so with this code they are 100% sure, without 99.99999% :)

The second question is simple:

Check how this code will work for ctx->total[0] == 0xFFFFFFFF and ilen == 1

ctx->total[0] += (uint32_t) ilen; // this will overflow and total[0] now 0

 if( ctx->total[0] < (uint32_t) ilen ) // 0<1 true ctx->total[1]++; 

Answer 2

if basically does:

 if (ctx->total[0] + (uint32_t) ilen < (uint32_t) ilen) ctx->total[1]++; 

i.e. if adding ilen to ctx->total[0] will overflow, then bump ctx->total[1] . In other words, it performs an add transfer.