C increment/decrement operators [closed] - c

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Determine the value of each variable after calculation is performed. All variables have value 5 before execution.
A1/=++B1/--C1
A2+=++B2%C2--
please tell me how this work

Answer 1

Variables:
int A1 = 5;
int B1 = 5;
int C1 = 5;
int A2 = 5;
int B2 = 5;
int C2 = 5;
Your code:
A1 /= ++B1 / --C1;
A2 += ++B2 % C2--;
Will probably compile into something similar too:
++B1;
--C1;
A1 /= B1 / C1;
++B2;
A2 += B2 % C2;
C2--;
You can output the ASM using your compiler, with GCC its the -S flag. Here is the ASM output with GCC on my computer (I added the comments):
movl $5, -20(%rbp) // A1 = 5
movl $5, -24(%rbp) // B1 = 5
movl $5, -28(%rbp) // C1 = 5
movl $5, -32(%rbp) // A2 = 5
movl $5, -36(%rbp) // B2 = 5
movl $5, -40(%rbp) // C2 = 5
Then for the first one calculation, this is performed (comments simplified for easier understanding):
addl $1, -24(%rbp) // ++B1
subl $1, -28(%rbp) // --C1
movl -24(%rbp), %eax //
cltd
idivl -28(%rbp) // divide B1 by C1
movl %eax, %esi //
movl -20(%rbp), %eax //
cltd
idivl %esi // divide A1 by the previous
movl %eax, -20(%rbp)

Answer 2

By the C Operator Precedence Table: http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm
For 1:
A1 /= ++B1 / --C1
C1 will first be decremented by 1 to 4
B1 will then be incremented by 1 to 6
B1 (6) will be divided by C1 (4), result of which will be 1
A1 will be assigned with the result of the division A1 (5) and 1, which is 5
Results for each will be 5, 6 and 4 for A1, B1 and C1, respectively.
For 2:
A2 += ++B2 % C2--
C2 will first be marked to get decremented close to the end of the statement, remains 5 for now
B2 will then be incremented by 1 to 6
The remainder from the division B2 (6) by C2 (5) will get calculated, which is 1
A2 will be assigned with the result of the addition of A2 (5) and 1, which is 6
C2 will be decremented by 1 to 4
Results for each will be 6, 6 and 4 for A2, B2 and C2, respectively.
Pardon me if I have made any mistakes, you can always check these with your compiler.