C Programming Questions - Part 2

1) How do you determine the size and range of the following data types ?

• char
• unsigned char
• short int
• unsigned int
• unsigned long float

Ans:-

limits.h header file defines the minimum and maximum range macros for each of the data types , sizeof(datatype) returns the number of bytes used by the datatype in current machine.

2) Write logical expressions that tests whether a given character variable c is

• lowercase letter
• uppercase letter
• digit
• white space(includes space,tab,newline)

Ans:-

• lowercase letter = (c >= 'a' && c <= 'z')
• uppercase letter = (c >= 'A' && c <= 'Z')
• digit = (c >= '0' && c <= '9')
• white space(includes space,tab,newline) = (c == ' ' || c == '\t' || c == '\n')

3) Consider unsigned int val=0xCAFE; Write expressions using bit wise operators that do the following:

• test if at least three of last four bits(LSB) are on
• reverse the byte order(i.e.,produce val=0xFECA)
• rotate four bits(i.e.,produce val=0xECAF)

Ans:-

• test if at least three of last four bits(LSB) are on – 
  obtain the last last four digits value and check with the possible values 0x07,0x0B,0x0D,0x0E,0x0F.
  
  
  Unsigned int bits = (val & 0x0F) or (val & ~((~0)<< 4)) – LSB 4 bits (bits == 0x07 || bits == 0x0B || bits >= 0x0D)

• reverse the byte order(i.e.,produce val=0xFECA) -
  
  
  unsigned int val = 0xCAFE;
  
  
  unsigned int res =0;
  
  
  res = ((val & 0xff ) << 8) | (val >> 8);
• rotate four bits(i.e.,produce val=0xECAF) - 
  unsigned int val = 0xCAFE;
  unsigned int res =0;
  res = ( val >> 4) | ((val & 0xf) << 12);

4) Using precedence rules,evaluate the following expressions and determine the value of the variables (without running the code).Also rewrite them using parenthesis to make the order explicit. 

• Assume(x=0xFF33,MASK=0xFF00).Expression:c=x&MASK==0;
• Assume(x=10,y=2,z=2;).Expression:z=y=x+++++y∗2;
• Assume(x=10,y=4,z=1;).Expression:y>>=x&0x2&&z

Ans:-

• The operator precedence is ==,&,=.
• Thus,the expression is equivalent to c=(x&(MASK==0)). 
• i,e x=0xFF33,c=0.

• The operator precedence is ++,*,+.
• Thus,the expression is equivalent to z=(x++)+((++y)∗2).
• i,e x=11,y=3,z=10+3∗2=16.

• The operator precedence is &,&&,>>=.
• Thus,the expression is equivalent to y>>=(x&0x2)&&z.
• i,e x=10,y=2,z=1.

5) Determine if the following statements have any errors. If so,highlight them and explain why.

•  int 2ndvalue=10;
• Assume(x=0,y=0,alliszero=1).alliszero=(x=1)&&(y=0);
• Assume(x=10,y=3,z=0;).y=++x+y;z=z−−>x;
• Assume that we want to test if last four bits of x are on. 
• (int MASK=0xF;ison=x&MASK==MASK)

Ans:-

• variable value should not start with digit.

•  = operator should be replaced with ==.i,e alliszero=(x==1)&&(y==0).

• this is confusing statement but its correct. y=(++x)+y;z=(z−−)>x;

• we need to add () to expression if we need the desired result 

• (int MASK=0xF;ison=x&(MASK==MASK))