Problem Set 2 — Solutions

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.087: Practical Programming in C IAP 2010 Problem Set 2 – Solutions Types, operators, expressions Out: Tuesday, January 12, 2010. Due: Wednesday, January 13, 2010. Problem 2.1 Determine the size,minimum and maximum value following data types. Please specify if your machine is 32 bit or 64 bits in the answer. • char • unsigned char • short • int • unsigned int • unsigned long • float Hint: Use sizeof() operator,limits.h and float.h header files Answer: On my 32-bit machine (/usr/include/limits.h,/usr/include/float.h), the sizes and limits are as follows. Results may differ if you used a 64 bit machine. Data type char unsigned char short int unsigned int unsigned long float size (bytes) 1 1 2 4 4 4 4 min SCHAR MIN (-128) 0 SHRT MIN (-32768) INT MIN (-2147483648) 0 0 FLT MIN(1.175494e-38) max SCHAR MAX (127) UCHAR MAX(255) SHRT MAX (32767) INT MAX (2147483647) UINT MAX(4294967295) ULONG MAX(4294967295) FLT MAX(3.402823e+38) Problem 2.2 Write logical expressions that tests whether a given character variable c is • lower case letter (Answer: c>=’a’&& c<=’z’) • upper case letter (Answer: c>=’A’&& c<=’Z’) • digit (Answer: c>=’0’&& c<=’9’) • white space (includes space,tab,new line) (Answer: c==’\n’|| c==’\t’|| c==’’) Problem 2.3 Consider int val=0xCAFE; Write expressions using bitwise operators that do the following: (a) test if atleast three of last four bits (LSB) are on (b) reverse the byte order (i.e., produce val=0xFECA) (c) rotate fourbits (i.e., produce val=0xECAF) Answer: (a) We have to test if last three or four bits are on. The possible values are 0x7,0xB,0xD,0xE,0xF. To test this, first we extract the last four bits. ( int bits=val&0xF; /∗last four bits∗/). Next we test if it is one of the possible patterns. ( bits==0x7 || bits==0xB || (bits>=0xD)). (b) val = ((0xFF & val) << 8) | (val>>8) (b) val = (val >> 4) | ((val&0xF)<<12) Problem 2.4 Using precedence rules, evaluate the following expressions and determine the value of the vari­ ables(without running the code). Also rewrite them using parenthesis to make the order explicit. (a) Assume (x=0xFF33,MASK=0xFF00).Expression: c=x & MASK ==0; (b) Assume (x=10,y=2,z=2;).Expression: z=y=x++ + ++y∗2; (c) Assume (x=10,y=4,z=1;).Expression: y>>= x&0x2 && z Answer: (a) The operator precedence is ’==’>’&’>’=’. Thus, the expression is equivalent to c= (x & (MASK==0)). Therefore x=0xFF33,c=0. (b) The operator precedence is ’++’>’*’>’+’. Thus, the expression is equivalent to z =(x++) + ((++y)∗2). Therefore x=11,y=3,z=10+3∗2=16. (b) The operator precedence is ’&’>’&&’>’>>=’. Thus, the expression is equivalent to y>>= (x & 0x2) && z. Therefore x=10,y=2,z=1. Problem 2.5 Determine if the following statements have any errors. If so, highlight them and explain why. (a) int 2nd value=10; (b) Assume (x=0,y=0,alliszero=1). alliszero =(x=1) && (y=0); (c) Assume (x=10,y=3,z=0;). y=++x+y;z=z−−>x; (d) Assume that we want to test if last four bits of x are on. (int MASK=0xF;ison=x&MASK==MASK) Answer: (a) Variable names cannot start with a number. (b) ’=’ operator should be replaced with ’==’. The correct version is alliszero =(x==1) && (y==0);. (c) There is nothing wrong with the statement. While −−> may look suspicious, the expression symplifies to y= (++x)+y; z=(z−−)>x. (c) There is nothing syntatically wrong with the statement. However, what we want is ison=(x&MASK)==MASK. Based on operator precedence, the current expression simplifies to ison=x&(MASK==MASK) These exercises should have convinced you to use () always.