post increment int single pointer

In this section, you are going to learn

How to iterate integer array using ptr++ ?

What is the meaning of ptr++, *ptr++, *(ptr++), *(ptr)++, (*ptr)++ ?

What is the meaning of c = ptr++, c = *ptr++, c = *(ptr++), c = *(ptr)++, c = (*ptr)++ ?

What is the difference between ptr++, *ptr++, *(ptr++), *(ptr)++, (*ptr)++ ?

What is the difference between c = ptr++, c = *ptr++, c = *(ptr++), c = *(ptr)++, c = (*ptr)++ ?

Topics of Post Increment

• Basics of Post Increment

• Meaning of expressions

• Summary of expressions

• Post Increment : int pointer inside structure

• Post Increment : Function Call

Basics of Post Increment

• ptr++ Basic Usage

• ptr++ : Iterate and print integers of an integer array

Example 1 : ptr++ : Basic Usage

• Step 1 : Define a Single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Point ptr to Single dimension array

int *ptr;

ptr = a;

• Step 3 : Increment ptr

ptr++;

• Step 4 : Print *ptr

printf("*ptr = %d\n", *ptr);

• See full program below

#include <stdio.h>

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        int *ptr;

        ptr = a;

        ptr++;

        printf("*ptr = %d\n", *ptr);

        return 0;
}

Example 2 : ptr++ : Iterate and print integers of an integer array

• Step 1 : Define a Single Dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Point ptr to Single dimension array

int *ptr;

ptr = a;

• Step 3 : Print integer array by iterating through all integers using ptr++

for (int i = 0; i < 7; i++)
{
        printf("*ptr = %d\n", *ptr);
        ptr++;
}

• See full program below

#include <stdio.h>

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        int *ptr;

        ptr = a;

        for (int i = 0; i < 7; i++)
        {
                printf("*ptr = %d\n", *ptr);
                ptr++;
        }

        return 0;
}

Meaning of expressions

• Meaning of ptr++ and c = ptr++

• Meaning of *(ptr++) and c = *(ptr++)

• Meaning of *ptr++ and c = *ptr++

• Meaning of *(ptr)++ and c = *(ptr)++

• Meaning of (*ptr)++ and c = (*ptr)++

Meaning of ptr++ and c = ptr++

• Consider statement

p = ptr++;

• There are two steps in this statement

  • Current value of ptr is assigned to p

  • ptr is incremented

• We now derived a rule

  • First Assign, then increment

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Define a single pointer and point to array

int *ptr;

ptr = a;

OR

int *ptr;

ptr = &a[0];

• Step 3 : Iterate and print the integer array

for (int i = 0; i < 7; i++)
{
        int *p;

        p = ptr++;

        printf("*p = %d\n", *p);
}

Output is as below

*p = 0
*p = 1
*p = 2
*p = 3
*p = 4
*p = 5
*p = 6

Can you guess what is happening ?

• Let us Recall

p = ptr++;

• There are two steps in this statement

  • Current value of ptr is assigned to p

  • ptr is incremented

• We now derived a rule

  • First Assign, then increment

• See full program below

#include <stdio.h>

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        int *ptr;

        ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int *p;

                p = ptr++;
                printf("*p = %d\n", *p);
        }

        return 0;
}

Meaning of *(ptr++) and c = *(ptr++)

• Consider statement

c = *(ptr++);

• There are two steps in this statement

  • Current value of *ptr is assigned to c

  • ptr is incremented

  • *ptr is NOT incremented

• We now derived a rule

  • First Assign, then increment

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Define a single pointer and point to array

int *ptr;

ptr = a;

OR

int *ptr;

ptr = &a[0];

• Step 3 : Iterate and print the integer array

for (int i = 0; i < 7; i++)
{
        int c;

        c = *(ptr++);

        printf("c = %d\n", c);
}

Output is as below

c = 0
c = 1
c = 2
c = 3
c = 4
c = 5
c = 6

Can you guess what is happening ?

• Let us Recall

c = *(ptr++);

• There are two steps in this statement

  • Current value of *ptr is assigned to c

  • ptr is incremented

  • *ptr is NOT incremented

• We now derived a rule

  • First Assign, then increment

• See full program below

#include <stdio.h>

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        int *ptr;

        ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *(ptr++);
                printf("c = %d\n", c);
        }

        return 0;
}

Meaning of *ptr++ and c = *ptr++

• Same as *(ptr++) and c = *(ptr++)

• Consider statement

c = *ptr++;

• There are two steps in this statement

  • Current value of *ptr is assigned to c

  • ptr is incremented

  • *ptr is NOT incremented

• We now derived a rule

  • First Assign, then increment

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Define a single pointer and point to array

int *ptr;

ptr = a;

OR

int *ptr;

ptr = &a[0];

• Step 3 : Iterate and print the integer array

for (int i = 0; i < 7; i++)
{
        int c;

        c = *ptr++;

        printf("c = %d\n", c);
}

Output is as below

c = 0
c = 1
c = 2
c = 3
c = 4
c = 5
c = 6

Can you guess what is happening ?

• Let us Recall

c = *ptr++;

• There are two steps in this statement

  • Current value of *ptr is assigned to c

  • ptr is incremented

  • *ptr is NOT incremented

• We now derived a rule

  • First Assign, then increment

• See full program below

#include <stdio.h>

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        int *ptr;

        ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *ptr++;
                printf("c = %d\n", c);
        }

        return 0;
}

Meaning of *(ptr)++ and c = *(ptr)++

• Same as *(ptr++) and c = *(ptr++)

• Consider statement

c = *(ptr)++;

• There are two steps in this statement

  • Current value of *ptr is assigned to c

  • ptr is incremented

  • *ptr is NOT incremented

• We now derived a rule

  • First Assign, then increment

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Define a single pointer and point to array

int *ptr;

ptr = a;

OR

int *ptr;

ptr = &a[0];

• Step 3 : Iterate and print the integer array

for (int i = 0; i < 7; i++)
{
        int c;

        c = *(ptr)++;

        printf("c = %d\n", c);
}

Output is as below

c = 0
c = 1
c = 2
c = 3
c = 4
c = 5
c = 6

Can you guess what is happening ?

• Let us Recall

c = *(ptr)++;

• There are two steps in this statement

  • Current value of *ptr is assigned to c

  • ptr is incremented

  • *ptr is NOT incremented

• We now derived a rule

  • First Assign, then increment

• See full program below

#include <stdio.h>

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        int *ptr;

        ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *(ptr)++;
                printf("c = %d\n", c);
        }

        return 0;
}

Meaning of (*ptr)++ and c = (*ptr)++

• Consider statement

c = (*ptr)++;

• There are two steps in this statement

  • Current value of *ptr is assigned to c

  • ptr is NOT incremented

  • *ptr is incremented

• We now derived a rule

  • First Assign, then increment

Summary of expressions

• Summary of Post Increment expressions

Summary of Post Increment expressions

Expression	Explanation
c = ptr++	Assign ptr to c Increment ptr
c = *(ptr++)	Assign *ptr to c Increment ptr DO NOT increment *ptr
c = *ptr++	Assign *ptr to c Increment ptr DO NOT increment *ptr
c = *(ptr)++	Assign *ptr to c Increment ptr DO NOT increment *ptr
c = (*ptr)++	Assign *ptr to c DO NOT increment ptr Increment *ptr

Post Increment : int pointer inside structure

• Integer pointer accessed using Structure Object

• Integer pointer accessed using Structure Single Pointer

• Integer pointer accessed using Structure Double Pointer

• Integer pointer accessed using Structure Triple Pointer

• Integer pointer accessed using Nested Structure Object

• Integer pointer accessed using Nested Structure Single Pointer

• Integer pointer accessed using Nested Structure 2 level Pointers

Integer pointer accessed using Structure Object

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Create a structure object

struct ABC abc;

• Step 3 : Point single integer pointer to single dimension array

abc.ptr = a;

• Step 4 : Iterate through single dimension array using pointer

for (int i = 0; i < 7; i++)
{
        int c;

        c = *abc.ptr++;
        printf("c = %d\n", c);
}

• See full program below

#include <stdio.h>

struct ABC
{
        int *ptr;
};

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        struct ABC abc;

        abc.ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *abc.ptr++;
                printf("c = %d\n", c);
        }

        return 0;
}

Integer pointer accessed using Structure Single Pointer

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Allocate memory for structure pointers

struct ABC *sp;

sp = malloc(sizeof(struct ABC));

• Step 3 : Point single integer pointer to single dimension array

sp->ptr = a;

• Step 4 : Iterate through single dimension array using pointer

for (int i = 0; i < 7; i++)
{
        int c;

        c = *sp->ptr++;
        printf("c = %d\n", c);
}

• Step 5 : Free memory after use

free(sp);

• See full program below

#include <stdio.h>
#include <stdlib.h>

struct ABC
{
        int *ptr;
};

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        struct ABC *sp;

        sp = malloc(sizeof(struct ABC));

        sp->ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *sp->ptr++;
                printf("c = %d\n", c);
        }

        free(sp);

        return 0;
}

Integer pointer accessed using Structure Double Pointer

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Allocate memory for structure pointers

struct ABC **dp;

dp = malloc(sizeof(struct ABC *));
*dp = malloc(sizeof(struct ABC ));

• Step 3 : Point single integer pointer to single dimension array

(*dp)->ptr = a;

• Step 4 : Iterate through single dimension array using pointer

for (int i = 0; i < 7; i++)
{
        int c;

        c = *(*dp)->ptr++;
        printf("c = %d\n", c);
}

• Step 5 : Free memory after use

free(*dp);
free(dp);

• See full program below

#include <stdio.h>
#include <stdlib.h>

struct ABC
{
        int *ptr;
};

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        struct ABC **dp;

        dp = malloc(sizeof(struct ABC *));
        *dp = malloc(sizeof(struct ABC ));

        (*dp)->ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *(*dp)->ptr++;
                printf("c = %d\n", c);
        }

        free(*dp);
        free(dp);

        return 0;
}

Integer pointer accessed using Structure Triple Pointer

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Allocate memory for structure pointers

struct ABC ***dp;

dp = malloc(sizeof(struct ABC **));
*dp = malloc(sizeof(struct ABC *));
**dp = malloc(sizeof(struct ABC ));

• Step 3 : Point single integer pointer to single dimension array

(**dp)->ptr = a;

• Step 4 : Iterate through single dimension array using pointer

for (int i = 0; i < 7; i++)
{
        int c;

        c = *(**dp)->ptr++;
        printf("c = %d\n", c);
}

• Step 5 : Free memory after use

free(**dp);
free(*dp);
free(dp);

• See full program below

#include <stdio.h>
#include <stdlib.h>

struct ABC
{
        int *ptr;
};

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        struct ABC ***dp;

        dp = malloc(sizeof(struct ABC **));
        *dp = malloc(sizeof(struct ABC *));
        **dp = malloc(sizeof(struct ABC ));

        (**dp)->ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *(**dp)->ptr++;
                printf("c = %d\n", c);
        }

        free(**dp);
        free(*dp);
        free(dp);

        return 0;
}

Integer pointer accessed using Nested Structure Object

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Create a structure object

struct ABC abc;

• Step 3 : Point single integer pointer to single dimension array

abc.pqr.ptr = a;

• Step 4 : Iterate through single dimension array using pointer

for (int i = 0; i < 7; i++)
{
        int c;

        c = *abc.pqr.ptr++;
        printf("c = %d\n", c);
}

• See full program below

#include <stdio.h>

struct PQR
{
        int *ptr;
};

struct ABC
{
        struct PQR pqr;
};

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        struct ABC abc;

        abc.pqr.ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *abc.pqr.ptr++;
                printf("c = %d\n", c);
        }

        return 0;
}

Integer pointer accessed using Nested Structure Single Pointer

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Allocate memory for structure pointers

struct ABC *sp;

sp = malloc(sizeof(struct ABC));

• Step 3 : Point single integer pointer to single dimension array

sp->pqr.ptr = a;

• Step 4 : Iterate through single dimension array using pointer

for (int i = 0; i < 7; i++)
{
        int c;

        c = *sp->pqr.ptr++;
        printf("c = %d\n", c);
}

• Step 5 : Free memory after use

free(sp);

• See full program below

#include <stdio.h>
#include <stdlib.h>

struct PQR
{
        int *ptr;
};

struct ABC
{
        struct PQR pqr;
};

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        struct ABC *sp;

        sp = malloc(sizeof(struct ABC));

        sp->pqr.ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *sp->pqr.ptr++;
                printf("c = %d\n", c);
        }

        free(sp);

        return 0;
}

Integer pointer accessed using Nested Structure 2 level Pointers

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Allocate memory for structure pointers

struct ABC *abc;

abc = malloc(sizeof(struct ABC));
abc->pqr = malloc(sizeof(struct PQR));

• Step 3 : Point single integer pointer to single dimension array

abc->pqr->ptr = a;

• Step 4 : Iterate through single dimension array using pointer

for (int i = 0; i < 7; i++)
{
        int c;

        c = *abc->pqr->ptr++;
        printf("c = %d\n", c);
}

• Step 5 : Free memory after use

free(abc->pqr);
free(abc);

• See full program below

#include <stdio.h>
#include <stdlib.h>

struct PQR
{
        int *ptr;
};

struct ABC
{
        struct PQR *pqr;
};

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        struct ABC *abc;

        abc = malloc(sizeof(struct ABC));
        abc->pqr = malloc(sizeof(struct PQR));

        abc->pqr->ptr = a;

        for (int i = 0; i < 7; i++)
        {
                int c;

                c = *abc->pqr->ptr++;
                printf("c = %d\n", c);
        }

        free(abc->pqr);
        free(abc);

        return 0;
}

Post Increment : Function Call

• *ptr++ : Function Call

• ptr++ : Function Call

*ptr++ : Function Call

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Define a single pointer and point to array

int *ptr;

ptr = a;

• Step 3 : Iterate array using ptr++ and Pass by Value

for (int i = 0; i < 7; i++)
{
        fun(*ptr++);
}

• Step 4 : Define a function fun which receives a integer from caller

void fun(int c)
{
        printf("c = %d\n", c);
}

Let us Recall,

• In case of *ptr++

  • Current value of *ptr is first assigned, then ptr incremented

• Hence, in this case

  • Current value of *ptr is first passed to function fun, then ptr incremented

• See full program below

#include <stdio.h>

void fun(int c)
{
        printf("c = %d\n", c);
}

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        int *ptr;

        ptr = a;

        for (int i = 0; i < 7; i++)
        {
                fun(*ptr++);
        }

        return 0;
}

ptr++ : Function Call

• Step 1 : Define a single dimension array

int a[7] = {0, 1, 2, 3, 4, 5, 6};

• Step 2 : Define a single pointer and point to array

int *ptr;

ptr = a;

• Step 3 : Iterate array using ptr++ and Pass by Value

for (int i = 0; i < 7; i++)
{
        fun(ptr++);
}

• Step 4 : Define a function fun which receives a integer pointer from caller

void fun(int *c)
{
        printf("*c = %d\n", *c);
}

Let us Recall,

• In case of ptr++

  • Current value of ptr is first assigned, then incremented

• Hence, in this case

  • Current value of ptr is first passed to function fun, then incremented

• See full program below

#include <stdio.h>

void fun(int *c)
{
        printf("*c = %d\n", *c);
}

int main(void)
{
        int a[7] = {0, 1, 2, 3, 4, 5, 6};

        int *ptr;

        ptr = a;

        for (int i = 0; i < 7; i++)
        {
                fun(ptr++);
        }

        return 0;
}

See Also

• Current Module

  • Post Increment with Pointers

• Previous Module

  • Pre Increment with Pointers

• Next Module

  • Pre decrement with Pointers

• Other Modules

  • Basic Equations

  • Basic Arrays

  • Malloc and Pointers

  • Type-Casting and Pointers

  • Functions and Pointers

  • Memcpy and Pointers

  • Const Pointers

  • void Pointers

  • Array of Pointers

  • Pointer to an Array

  • Post Decrement with Pointers