Sequential Search array for loop condition












0















If length of array in sequential search is not given and there is a lot of out of order indexing,




  1. how many times we should execute the for loop?

  2. what should be the condition of for loop?


I counted indexing again and again. 



#include <iostream>

using namespace std;

int main() {

  int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                 300, 400, 5006, 110, 550, 440, 330, 331, 41};



  // length of array is not provided



  int i, n, loc = -1;

  cout << "Enter value to find" << endl;

  cin >> n;

  for (i = 0; i < 5; i++) {

    if (array[i] == n) {

      loc = i;

    }

  }

  if (loc == -1)

    cout << "Number is not found in array" << endl;

  else

    cout << "the number is found at position " << loc << " and is " << n;

  return 0;

}


For loop Condition for unknown length array in sequential search






c++ arrays







share|improve this question




















  • 1





    for (i = 0; i < sizeof array / sizeof *array; i++)

    – Sid S
    Jan 2 at 4:50






  • 5





    Possible duplicate of How do I find the length of an array?

    – user10605163
    Jan 2 at 4:50











  • Arrays never have an unknown length, all arrays will have their size fixed at compilation time. There is a (relatively) easy "trick" to get the size of an array (a proper array, not one that has decayed to a pointer): Dividing the size of the whole array with the size of a single element. Like (in your case) sizeof array / sizeof array[0].

    – Some programmer dude
    Jan 2 at 4:51











  • main must have a return type and it must be int.

    – user10605163
    Jan 2 at 4:51
















0















If length of array in sequential search is not given and there is a lot of out of order indexing,




  1. how many times we should execute the for loop?

  2. what should be the condition of for loop?


I counted indexing again and again. 



#include <iostream>

using namespace std;

int main() {

  int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                 300, 400, 5006, 110, 550, 440, 330, 331, 41};



  // length of array is not provided



  int i, n, loc = -1;

  cout << "Enter value to find" << endl;

  cin >> n;

  for (i = 0; i < 5; i++) {

    if (array[i] == n) {

      loc = i;

    }

  }

  if (loc == -1)

    cout << "Number is not found in array" << endl;

  else

    cout << "the number is found at position " << loc << " and is " << n;

  return 0;

}


For loop Condition for unknown length array in sequential search






c++ arrays







share|improve this question




















  • 1





    for (i = 0; i < sizeof array / sizeof *array; i++)

    – Sid S
    Jan 2 at 4:50






  • 5





    Possible duplicate of How do I find the length of an array?

    – user10605163
    Jan 2 at 4:50











  • Arrays never have an unknown length, all arrays will have their size fixed at compilation time. There is a (relatively) easy "trick" to get the size of an array (a proper array, not one that has decayed to a pointer): Dividing the size of the whole array with the size of a single element. Like (in your case) sizeof array / sizeof array[0].

    – Some programmer dude
    Jan 2 at 4:51











  • main must have a return type and it must be int.

    – user10605163
    Jan 2 at 4:51














0












0








0








If length of array in sequential search is not given and there is a lot of out of order indexing,




  1. how many times we should execute the for loop?

  2. what should be the condition of for loop?


I counted indexing again and again. 



#include <iostream>

using namespace std;

int main() {

  int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                 300, 400, 5006, 110, 550, 440, 330, 331, 41};



  // length of array is not provided



  int i, n, loc = -1;

  cout << "Enter value to find" << endl;

  cin >> n;

  for (i = 0; i < 5; i++) {

    if (array[i] == n) {

      loc = i;

    }

  }

  if (loc == -1)

    cout << "Number is not found in array" << endl;

  else

    cout << "the number is found at position " << loc << " and is " << n;

  return 0;

}


For loop Condition for unknown length array in sequential search






c++ arrays







share|improve this question
















If length of array in sequential search is not given and there is a lot of out of order indexing,




  1. how many times we should execute the for loop?

  2. what should be the condition of for loop?


I counted indexing again and again. 



#include <iostream>

using namespace std;

int main() {

  int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                 300, 400, 5006, 110, 550, 440, 330, 331, 41};



  // length of array is not provided



  int i, n, loc = -1;

  cout << "Enter value to find" << endl;

  cin >> n;

  for (i = 0; i < 5; i++) {

    if (array[i] == n) {

      loc = i;

    }

  }

  if (loc == -1)

    cout << "Number is not found in array" << endl;

  else

    cout << "the number is found at position " << loc << " and is " << n;

  return 0;

}


For loop Condition for unknown length array in sequential search






c++ arrays




c++ arrays






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 2 at 6:27









William Miller

1,405217




1,405217










asked Jan 2 at 4:47









Sami UllahSami Ullah

11




11








  • 1





    for (i = 0; i < sizeof array / sizeof *array; i++)

    – Sid S
    Jan 2 at 4:50






  • 5





    Possible duplicate of How do I find the length of an array?

    – user10605163
    Jan 2 at 4:50











  • Arrays never have an unknown length, all arrays will have their size fixed at compilation time. There is a (relatively) easy "trick" to get the size of an array (a proper array, not one that has decayed to a pointer): Dividing the size of the whole array with the size of a single element. Like (in your case) sizeof array / sizeof array[0].

    – Some programmer dude
    Jan 2 at 4:51











  • main must have a return type and it must be int.

    – user10605163
    Jan 2 at 4:51














  • 1





    for (i = 0; i < sizeof array / sizeof *array; i++)

    – Sid S
    Jan 2 at 4:50






  • 5





    Possible duplicate of How do I find the length of an array?

    – user10605163
    Jan 2 at 4:50











  • Arrays never have an unknown length, all arrays will have their size fixed at compilation time. There is a (relatively) easy "trick" to get the size of an array (a proper array, not one that has decayed to a pointer): Dividing the size of the whole array with the size of a single element. Like (in your case) sizeof array / sizeof array[0].

    – Some programmer dude
    Jan 2 at 4:51











  • main must have a return type and it must be int.

    – user10605163
    Jan 2 at 4:51








1




1





for (i = 0; i < sizeof array / sizeof *array; i++)

– Sid S
Jan 2 at 4:50





for (i = 0; i < sizeof array / sizeof *array; i++)

– Sid S
Jan 2 at 4:50




5




5





Possible duplicate of How do I find the length of an array?

– user10605163
Jan 2 at 4:50





Possible duplicate of How do I find the length of an array?

– user10605163
Jan 2 at 4:50













Arrays never have an unknown length, all arrays will have their size fixed at compilation time. There is a (relatively) easy "trick" to get the size of an array (a proper array, not one that has decayed to a pointer): Dividing the size of the whole array with the size of a single element. Like (in your case) sizeof array / sizeof array[0].

– Some programmer dude
Jan 2 at 4:51





Arrays never have an unknown length, all arrays will have their size fixed at compilation time. There is a (relatively) easy "trick" to get the size of an array (a proper array, not one that has decayed to a pointer): Dividing the size of the whole array with the size of a single element. Like (in your case) sizeof array / sizeof array[0].

– Some programmer dude
Jan 2 at 4:51













main must have a return type and it must be int.

– user10605163
Jan 2 at 4:51





main must have a return type and it must be int.

– user10605163
Jan 2 at 4:51












1 Answer
1






active

oldest

votes


















0














Using sizeof(array)/sizeof(array[0]) will give you the number of elements in the array, 



#include <iostream>

using namespace std;

int main() {

    int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                   300, 400, 5006, 110, 550, 440, 330, 331, 41};



    // length of array is not provided



    int i, n, loc = -1;

    cout << "Enter value to find" << endl;

    cin >> n;

    for (i = 0; i < sizeof(array) / sizeof(array[0]); i++) {

        if (array[i] == n) {

            loc = i;

        }

    }

    if (loc == -1)

    cout << "Number is not found in array" << endl;

    else

    cout << "the number is found at position " << loc << " and is " << n;

    return 0;

}


Note that using namespace std; is considered bad practice`






share|improve this answer


























  • "using namespace std; is considered bad practice" That's still an opinion.

    – Sid S
    Jan 2 at 5:43













  • @SidS you're correct. However it is not my opinion - I am simply relaying it

    – William Miller
    Jan 2 at 5:45











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1 Answer
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1






active

oldest

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oldest

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active

oldest

votes









0














Using sizeof(array)/sizeof(array[0]) will give you the number of elements in the array, 



#include <iostream>

using namespace std;

int main() {

    int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                   300, 400, 5006, 110, 550, 440, 330, 331, 41};



    // length of array is not provided



    int i, n, loc = -1;

    cout << "Enter value to find" << endl;

    cin >> n;

    for (i = 0; i < sizeof(array) / sizeof(array[0]); i++) {

        if (array[i] == n) {

            loc = i;

        }

    }

    if (loc == -1)

    cout << "Number is not found in array" << endl;

    else

    cout << "the number is found at position " << loc << " and is " << n;

    return 0;

}


Note that using namespace std; is considered bad practice`






share|improve this answer


























  • "using namespace std; is considered bad practice" That's still an opinion.

    – Sid S
    Jan 2 at 5:43













  • @SidS you're correct. However it is not my opinion - I am simply relaying it

    – William Miller
    Jan 2 at 5:45
















0














Using sizeof(array)/sizeof(array[0]) will give you the number of elements in the array, 



#include <iostream>

using namespace std;

int main() {

    int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                   300, 400, 5006, 110, 550, 440, 330, 331, 41};



    // length of array is not provided



    int i, n, loc = -1;

    cout << "Enter value to find" << endl;

    cin >> n;

    for (i = 0; i < sizeof(array) / sizeof(array[0]); i++) {

        if (array[i] == n) {

            loc = i;

        }

    }

    if (loc == -1)

    cout << "Number is not found in array" << endl;

    else

    cout << "the number is found at position " << loc << " and is " << n;

    return 0;

}


Note that using namespace std; is considered bad practice`






share|improve this answer


























  • "using namespace std; is considered bad practice" That's still an opinion.

    – Sid S
    Jan 2 at 5:43













  • @SidS you're correct. However it is not my opinion - I am simply relaying it

    – William Miller
    Jan 2 at 5:45














0












0








0







Using sizeof(array)/sizeof(array[0]) will give you the number of elements in the array, 



#include <iostream>

using namespace std;

int main() {

    int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                   300, 400, 5006, 110, 550, 440, 330, 331, 41};



    // length of array is not provided



    int i, n, loc = -1;

    cout << "Enter value to find" << endl;

    cin >> n;

    for (i = 0; i < sizeof(array) / sizeof(array[0]); i++) {

        if (array[i] == n) {

            loc = i;

        }

    }

    if (loc == -1)

    cout << "Number is not found in array" << endl;

    else

    cout << "the number is found at position " << loc << " and is " << n;

    return 0;

}


Note that using namespace std; is considered bad practice`






share|improve this answer















Using sizeof(array)/sizeof(array[0]) will give you the number of elements in the array, 



#include <iostream>

using namespace std;

int main() {

    int array = {10,  20, 30, 40, 50, 60, 70, 80, 90, 100, 200,

                   300, 400, 5006, 110, 550, 440, 330, 331, 41};



    // length of array is not provided



    int i, n, loc = -1;

    cout << "Enter value to find" << endl;

    cin >> n;

    for (i = 0; i < sizeof(array) / sizeof(array[0]); i++) {

        if (array[i] == n) {

            loc = i;

        }

    }

    if (loc == -1)

    cout << "Number is not found in array" << endl;

    else

    cout << "the number is found at position " << loc << " and is " << n;

    return 0;

}


Note that using namespace std; is considered bad practice`







share|improve this answer














share|improve this answer



share|improve this answer








edited Jan 2 at 5:44

























answered Jan 2 at 5:12









William MillerWilliam Miller

1,405217




1,405217













  • "using namespace std; is considered bad practice" That's still an opinion.

    – Sid S
    Jan 2 at 5:43













  • @SidS you're correct. However it is not my opinion - I am simply relaying it

    – William Miller
    Jan 2 at 5:45



















  • "using namespace std; is considered bad practice" That's still an opinion.

    – Sid S
    Jan 2 at 5:43













  • @SidS you're correct. However it is not my opinion - I am simply relaying it

    – William Miller
    Jan 2 at 5:45

















"using namespace std; is considered bad practice" That's still an opinion.

– Sid S
Jan 2 at 5:43







"using namespace std; is considered bad practice" That's still an opinion.

– Sid S
Jan 2 at 5:43















@SidS you're correct. However it is not my opinion - I am simply relaying it

– William Miller
Jan 2 at 5:45





@SidS you're correct. However it is not my opinion - I am simply relaying it

– William Miller
Jan 2 at 5:45




















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