Final value of ch is___ (in hex) unsigned char my char = 0x76; ch = (ch 3); Fina

Question

Final value of ch is___ (in hex) unsigned char my char = 0x76; ch = (ch 3); Final value of ptr char is ___ (in hex) unsigned char my_char = 0x8F;//Assume mu char is located at char *ptr char; my char = my char/8; ptr char = &my; char + my char; Final value of char ptr is ___ (in hex) unsigned char val1 = 0xFF;//Assume val1 is located at 0xFFFF FFFE unsigned char val2 = 1;//Assume val2 is located at 0xFFFF FFF0 char *char ptr; if (val1 + val) {char ptr = & val1;} else {char ptr = &val2; Final value of char ch is___ (in decimal) unsigned char ch;//Assume ch is located at 0xFFFF FF00 unsigned char *ch_ptr = 0xFFFF_AAAA; ch = (*(ch_ptr++) + 1)

Explanation / Answer

(a) The value assigned to the character is 0x76 which has a value of 118 (7*16 + 6) when converted to decimal. When we perform the left shift operation (ch<<4) we get the value of 96 (011100000 in binary) and in the same way we get the value 14 (00001110) when we perform right shift (ch>>3). On performing ch=(ch<<4) | (ch>>3), we get the value 110(01101110) which is the final value. So, the final value of ch is 0x110

(b) The value assigned to the unsigned character my_char is 0x8F which is 143 (8*16 + 15) in decimal. When we perform 143/8 we get 17, so the value stored in my_char will be 17. The address of my_char is 0xFFFFFD0C initially, so when we compute ptr_char = (&my_char) + my_char, so we are moving 17 addresses ahead of 0xFFFFFD0C. This means that ptr_char will get the value 0xFFFFFD1D.

(c) We assign the value 0xFF to the first unsigned character val1, so the value stored will be 255(15*16 +15 =255) in decimal i.e 11111111 in binary. The second unsigned character val2 gets the value of 1. When we add the two values in binary we get 100000000. This number has been type cast as a unsigned character in the if statement. So, only the first 8 bits from the right are considered i.e 00000000. So, the if condition is evaluated to false. The else block will be executed and the value of the pointer char_ptr will be equal to the address of val2 i.e 0xFFFFFFF0.

(d) This question appears to be similar to part(c) but there is one small change which completely changes the entire answer. We assign the value 0xFF to the first unsigned character val1, so the value stored will be 255(15*16 +15 =255) in decimal i.e 11111111 in binary. The second unsigned character val2 gets the value of 1. When we calculate val1 + val2, we get 100000000 in binary. This is given as the condition for the if statement but there isnt any type casting here so, since this value is greater than 0, it represents a true case, so the statements in the if block will be executed. So the character pointer char_ptr will have the address of val1 i.e 0xFFFFFFFE.

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