Tutorial 1: C-Language
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1 Tutorial 1: C-Language Problem 1: Data Type What are the ranges of the following data types? int 32 bits OR ( if unsiged) in some machines int is same as short long same as int long long 64 bits OR ( if unsigned) short 16 bits OR ( if unsigned) char 8 bits OR ( if unsigned) Problem 2: Data Type Conversion Consider the C code in Figure 1. Answer the following questions. What are the meaning of statements (b = (short) a;) and (c = (char) a;)? What are the outputs of the printf statements if (a = )? short b; char c; b = (short) a; c = (char) a; printf("short = \"%d\"\n\n", b); printf("char = \"%d\"\n\n", c); Figure 1: Program on Data Type Conversion Two statements (b = (short) a;) and (c = (char) a;) are called type casts. They convert one data type to another. In order to understand how they work and what are printed out we must start with representing (a = ) in hex (base 16) format. We can do this using a calculator. In hex format (a = ) is represented as (a = 0xf ). Two statements (b = (short) a;) and (c = (char) a;) have the following effects: (b = (short) a;) : (a = 0xf ) in 32 bits (b = 0x2222) in 16 bits (c = (char) a;) : (a = 0xf ) in 32 bits (c = 0x22) in 8 bits Converting back from the hex format to decimal give (b = 8738) and (c = 34). 1
2 integer = " " short = "8738" char = "34" Figure 2: The Outputs of printf Statements For (a = ) the hex representation is (a = 0xf222f2f2). Two statements (b = (short) a;) and (c = (char) a;) have the following effects: (b = (short) a;) : (a = 0xf222f2f2) in 32 bits (b = 0xf2f2) in 16 bits (c = (char) a;) : (a = 0xf222f2f2) in 32 bits (c = 0xf2) in 8 bits Converting back from the hex format to decimal give (b = -3342) and (c = -14). integer = " " short = "-3342" char = "-14" Figure 3: The Outputs of printf Statements Problem 3: More Data Type Conversion Consider the C code in Figure 4. Answer the following questions. What are the meaning of statements (a = (int) b;) and (c = (char) b;)? What are the outputs of the printf statements if (b = )? short b= ; int a; char c; a = (int) b; c = (char) b; printf("short = \"%d\"\n\n", b); printf("char = \"%d\"\n\n", c); Figure 4: Program on Data Type Conversion Let us see how the two type casts in statements (b = (short) a;) and (c = (char) a;) work. In order to understand how they work and what are printed let us represent (b = ) in hex format as (b = 0x80000). Two statements (a = (int) b;) and (c = (char) b;) have the following effects: (a = (int) b;) : (b = 0x8000) in 16 bits (a = 0xffff8000) in 32 bits (c = (char) b;) : (b = 0x8000) in 16 bits (c = 0x00) in 8 bits Converting back from the hex format to decimal give (a = ) and (c = 0). 2
3 short = "-32768" integer = "-32768" char = "0" Figure 5: The Outputs of printf Statements For (b = ) the hex representation is (a = 0x8780). Two statements (a = (int) b;) and (c = (char) b;) have the following effects: (a = (int) b;) : (b = 0x8780) in 16 bits (a = 0xffff8780) in 32 bits (c = (char) b;) : (b = 0x8780) in 16 bits (c = 0x80) in 8 bits Converting back from the hex format to decimal give (a = ) and (c = -128). short = "-30848" integer = "-30848" char = "-128" Figure 6: The Outputs of printf Statements Problem 4: Some More Data Type Conversion Consider the C code in Figure 7. Answer the following questions. What are the meaning of statements (b = (int)(short) a;) and (c = (int)(char) a;)? What are the outputs of the printf statements if (a = )? int b; int c; b = (int)(short) a; c = (int)(char) a; printf("int_short = \"%d\"\n\n", b); printf("int_char = \"%d\"\n\n", c); Figure 7: Program on Data Type Conversion Two type casts statements (b = (int)(short) a;) and (c = (int)(char) a;) convert (a) to types (short) and (char) and back to (int). Again representing (a = ) as (a = 0xf ). Two statements (b = (int) (short) a;) and (c = (int) (char) a;) have the following effects: (b = (int)(short) a;) : (a = 0xf ) in 32 bits (0x2222) in 16 bits (0x2222) in 16 bits (b = 0x ) in 32 bits 3
4 (c = (int)(char) a;) : (a = 0xf ) in 32 bits (0x22) in 8 bits (0x22) in 8 bits (c = 0x ) in 32 bits Converting back from the hex format to decimal give (b = 8738) and (c = 34). integer = " " int_short = "8738" int_char = "34" Figure 8: The Outputs of printf Statements For (a = ) the hex representation is (a = 0xf222f2f2). Two statements (b = (int)(short) a;) and (c = (int)(char) a;) have the following effects: (b = (int)(short) a;) : (a = 0xf222f2f2) in 32 bits (0xf2f2) in 16 bits (0xf2f2) in 16 bits (b = 0xfffff2f2) in 32 bits (c = (char) a;) : (a = 0xf222f2f2) in 32 bits (0xf2) in 8 bits (0xf2) in 8 bits (c = 0xfffffff2) in 32 bits Converting back from the hex format to decimal give (b = -3342) and (c = -14). integer = " " int_short = "-3342" int_char = "-14" Figure 9: The Outputs of printf Statements Problem 5: Still Some More Data Type Conversion Consider the C code in Figure 10. Answer the following questions. What is the output of the last printf statement, if (+ 1) is removed from the statement (d =(short *) &a + 1;)? char *b, *c; short *d; b =(char *) &a; c =(char *) &a + 1; d =(short *) &a + 1; printf("char1 = \"%d\"\n\n", *b); printf("char2 = \"%d\"\n\n", *c); printf("short = \"%d\"\n\n", *d); Figure 10: Program on Data Type Conversion with Pointers 4
5 For (a = ) the hex representation is (a = 0xf ). We can represent a as contacanation of 4 bytes as: a = 0xf The statement (b = (char *) &a;) obtains the address of the int type variable a and converts is to the address of the char type through type cast (char *) and assigns it to pointer c of type char. So what gets printed out by the second printf statement the first (right most) byte of a which is 0x22 = 34. The (c = (char *) &a + 1;) will access the second (from right) byte of a which is again 0x22 = 34. The statement (d = (short *) &a +1;) obtains the address of the integer type variable a and converts is to the address of the short (16-bit) type through type cast (short *), addes 1 to it, and assigns it to pointer d of type short. So what gets printed out by the last printf statement the second (left) half of a which is 0xf222 = char1 = "34" char2 = "34" short = "-3550" Figure 11: The Outputs of printf Statements By dropping (+ 1) from the statement (d = (short *) &a + 1;) we obtains the address of first (right) half of a which is 0x2222 = So what gets printed out by the last printf statement is the first (right) half of a which is 0x2222 =
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