Fundamentals of Programming Session 19 Instructor: Reza - - PowerPoint PPT Presentation
Fundamentals of Programming Session 19 Instructor: Reza Entezari-Maleki Email: entezari@ce.sharif.edu 1 Fall 2013 These slides have been created using Deitels slides Sharif University of Technology Outlines Using the const Qualifier with
Fall 2013
Sharif University of Technology
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Session 19
These slides have been created using Deitel’s slides
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The const qualifier enables you to inform the compiler that
If an attempt is made to modify a value that is declared
There are four ways to pass a pointer to a function: a non-
Each of the four combinations provides different access
These are discussed in the next several examples.
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The highest level of data access is granted by a non-
In this case, the data can be modified through the
Function convertToUppercase of Fig. 7.10
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A non-constant pointer to constant data can be
Such a pointer might be used to receive an array
For
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Figure 7.12 illustrates the attempt to compile a
This function attempts to modify the data pointed to
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A constant pointer to non-constant data always
This is the default for an array name. Pointers
Figure 7.13 attempts to modify a constant pointer.
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The least access privilege is granted by a constant
Such a pointer always points to the same memory
This is how an array should be passed to a function that
Figure 7.14 defines pointer variable ptr (line 13) to be
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What will be the output of the following program?
int main() { char *p; p="hello"; printf("%s\n", *&*&p); return 0; }
Answer:
hello
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What will be the output of the following program?
int main() { printf("%c\n", 5["Salaam"]); return 0; }
Answer:
m
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What will be the output of the following program?
int main() { char *str; str = "A%d\n"; str++; str++; printf(str-1, 300); return 0; }
Answer:
300
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What will be the output of the following program?
int main() { int arr[2][2][2] = {10, 2, 3, 4, 5, 6, 7, 8}; int *p, *q; p = &arr[1][1][1]; q = arr[0][1]; printf("%d, %d\n", *p, *q); return 0; }
Answer:
8, 3
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C provides the special unary operator sizeof to
When applied to the name of an array as in Fig. 7.16
Variables of type float are normally stored in 4 bytes
Therefore, there are a total of 80 bytes in array.
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The number of elements in an array also can be determined
For example, consider the following array definition:
doubl
uble real[ eal[ 22 22 ]; ];
Variables of type double normally are stored in 8 bytes of
Thus, array real contains a total of 176 bytes. To determine the number of elements in the array, the
si
sizeo eof( ( real eal ) / / si sizeof eof( re ( real[ l[ 0 ] ) ] )
Figure 7.17 calculates the number of bytes used to store each
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Operator sizeof can be applied to any variable name, type or
value (including the value of an expression).
When applied to a variable name (that is not an array name) or
a constant, the number of bytes used to store the specific type
The parentheses used with sizeof are required if a type name
with two words is supplied as its operand (such as long double or unsigned short).
Omitting the parentheses in this case results in a syntax error. The parentheses are not required if a variable name or a one-
word type name is supplied as its operand, but they can still be included without causing an error.
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What will be the output of the program assuming that the array
int main() { int arr[2][2][2] = {10, 2, 3, 4, 5, 6, 7, 8}; int *p, *q; p = &arr[1][1][1]; q = (int*) arr; printf("%d, %d\n", *p, *q); return 0; }
Answer:
8 10
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What will be the output of the following program?
int main(){ register int a = 25; int *p; p=&a; printf("%d ",*p); return 0; }
Answer:
It depends! Basically it should generate an error.
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What will be the output of the following program?
int power(int**); int main() { int a=5, *aa; /* Address of 'a' is 1000 */ aa = &a; a = power(&aa); printf("%d\n", a); return 0; } int power(int **ptr) { int b; b = **ptr***ptr; return (b); }
Answer:
25
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What does the following code do in C?
int main() { char str[80]; char token[10]; char *p , *q; printf("Enter a sentence: "); scanf("%s",str); p = str; while (*p) { q = token; while (*p) { *q = *p; q++ ; p++; } if (*p) p++; *q = '\0'; printf("%s\n" , token); } return 0; }
Answer: It gets a sentence and prints it!
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What will be the output of the following program?
int main() { printf("%d, %d\n", sizeof(NULL), sizeof("")); return 0; }
Answer:
4, 1
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What will be the output of the following program?
int main() { int *p1,**p2; double *q1,**q2; printf("%d %d ",sizeof(p1),sizeof(p2)); printf("%d %d",sizeof(q1),sizeof(q2)); return 0; }
Answer:
4 4 4 4
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