Maximum call stack size error in running a simple recursive function
I'm pretty new to JavaScript programming but I had experience in so many other programming languages. The idea of recursive functions in JavaScript is a fresh subject for me and I didn't see anything similar in other languages that I worked with. So for the sake of practicing, I decided to write some of the programs that I already have written with "for loops".
One of these programs is a function that takes a string as its argument and reports how many B letters are in it. Using objective-oriented programming, I first declared a function that can find the number of any other characters within a string. The program is as follows,
function countChar(string, char) {
let counted = 0;
for (let index = 0; index < string.length; index++) {
if (string[index].toUpperCase() == char.toUpperCase()) {
counted += 1;
}
}
return counted;
}
function countBs(text) {
return countChar(text, 'B');
}
console.log(countBs('Baby'));
// output = 2
It works very well but now that I'm using recursive functions, I get the "Maximum call stack size" error. My program with recursive functions looks like this one,
function countChar(string, char) {
function cursor(i, counted) {
if (i == string.length) {
return counted;
} else if (string[i].toUpperCase() == char.toUpperCase()) {
return cursor(i++, counted++);
} else {
return cursor(i++, counted);
}
}
return cursor(0,0);
}
function countBs(text) {
return countChar(text, 'B');
}
console.log(countBs('Baby'));
// output must be 2 but I get 'Maximum call stack size' error instead :(
Can anybody offers a modification to this program in order to get the solution? Is it basically possible to write this program by using the recursive functions?
javascript
asked Nov 14 '18 at 16:05
Saeed AhadianSaeed Ahadian
1115
add a comment |
I'm pretty new to JavaScript programming but I had experience in so many other programming languages. The idea of recursive functions in JavaScript is a fresh subject for me and I didn't see anything similar in other languages that I worked with. So for the sake of practicing, I decided to write some of the programs that I already have written with "for loops".
One of these programs is a function that takes a string as its argument and reports how many B letters are in it. Using objective-oriented programming, I first declared a function that can find the number of any other characters within a string. The program is as follows,
function countChar(string, char) {
let counted = 0;
for (let index = 0; index < string.length; index++) {
if (string[index].toUpperCase() == char.toUpperCase()) {
counted += 1;
}
}
return counted;
}
function countBs(text) {
return countChar(text, 'B');
}
console.log(countBs('Baby'));
// output = 2
It works very well but now that I'm using recursive functions, I get the "Maximum call stack size" error. My program with recursive functions looks like this one,
function countChar(string, char) {
function cursor(i, counted) {
if (i == string.length) {
return counted;
} else if (string[i].toUpperCase() == char.toUpperCase()) {
return cursor(i++, counted++);
} else {
return cursor(i++, counted);
}
}
return cursor(0,0);
}
function countBs(text) {
return countChar(text, 'B');
}
console.log(countBs('Baby'));
// output must be 2 but I get 'Maximum call stack size' error instead :(
Can anybody offers a modification to this program in order to get the solution? Is it basically possible to write this program by using the recursive functions?
javascript
asked Nov 14 '18 at 16:05
Saeed AhadianSaeed Ahadian
1115
1
Consider renamingstringandcharto some not-so-system names.
– Justinas
Nov 14 '18 at 16:08
add a comment |
I'm pretty new to JavaScript programming but I had experience in so many other programming languages. The idea of recursive functions in JavaScript is a fresh subject for me and I didn't see anything similar in other languages that I worked with. So for the sake of practicing, I decided to write some of the programs that I already have written with "for loops".
One of these programs is a function that takes a string as its argument and reports how many B letters are in it. Using objective-oriented programming, I first declared a function that can find the number of any other characters within a string. The program is as follows,
function countChar(string, char) {
let counted = 0;
for (let index = 0; index < string.length; index++) {
if (string[index].toUpperCase() == char.toUpperCase()) {
counted += 1;
}
}
return counted;
}
function countBs(text) {
return countChar(text, 'B');
}
console.log(countBs('Baby'));
// output = 2
It works very well but now that I'm using recursive functions, I get the "Maximum call stack size" error. My program with recursive functions looks like this one,
function countChar(string, char) {
function cursor(i, counted) {
if (i == string.length) {
return counted;
} else if (string[i].toUpperCase() == char.toUpperCase()) {
return cursor(i++, counted++);
} else {
return cursor(i++, counted);
}
}
return cursor(0,0);
}
function countBs(text) {
return countChar(text, 'B');
}
console.log(countBs('Baby'));
// output must be 2 but I get 'Maximum call stack size' error instead :(
Can anybody offers a modification to this program in order to get the solution? Is it basically possible to write this program by using the recursive functions?
javascript
asked Nov 14 '18 at 16:05
Saeed AhadianSaeed Ahadian
1115
I'm pretty new to JavaScript programming but I had experience in so many other programming languages. The idea of recursive functions in JavaScript is a fresh subject for me and I didn't see anything similar in other languages that I worked with. So for the sake of practicing, I decided to write some of the programs that I already have written with "for loops".
One of these programs is a function that takes a string as its argument and reports how many B letters are in it. Using objective-oriented programming, I first declared a function that can find the number of any other characters within a string. The program is as follows,
function countChar(string, char) {
let counted = 0;
for (let index = 0; index < string.length; index++) {
if (string[index].toUpperCase() == char.toUpperCase()) {
counted += 1;
}
}
return counted;
}
function countBs(text) {
return countChar(text, 'B');
}
console.log(countBs('Baby'));
// output = 2
It works very well but now that I'm using recursive functions, I get the "Maximum call stack size" error. My program with recursive functions looks like this one,
function countChar(string, char) {
function cursor(i, counted) {
if (i == string.length) {
return counted;
} else if (string[i].toUpperCase() == char.toUpperCase()) {
return cursor(i++, counted++);
} else {
return cursor(i++, counted);
}
}
return cursor(0,0);
}
function countBs(text) {
return countChar(text, 'B');
}
console.log(countBs('Baby'));
// output must be 2 but I get 'Maximum call stack size' error instead :(
Can anybody offers a modification to this program in order to get the solution? Is it basically possible to write this program by using the recursive functions?
javascript
javascript
asked Nov 14 '18 at 16:05
Saeed AhadianSaeed Ahadian
1115
asked Nov 14 '18 at 16:05
Saeed AhadianSaeed Ahadian
1115
asked Nov 14 '18 at 16:05
Saeed AhadianSaeed Ahadian
1115
asked Nov 14 '18 at 16:05
Saeed AhadianSaeed Ahadian
1115
asked Nov 14 '18 at 16:05
Saeed AhadianSaeed Ahadian
1115
1115
1
Consider renamingstringandcharto some not-so-system names.
– Justinas
Nov 14 '18 at 16:08
add a comment |
1
Consider renamingstringandcharto some not-so-system names.
– Justinas
Nov 14 '18 at 16:08
1
1
Consider renaming
string and char to some not-so-system names.– Justinas
Nov 14 '18 at 16:08
Consider renaming
string and char to some not-so-system names.– Justinas
Nov 14 '18 at 16:08
add a comment |
1 Answer
1
active
oldest
votes
because
return cursor(i++, counted++);
has to be
return cursor(i + 1, counted + 1);
(as you want to increase the value passed recursively, not the local variable i)
How i would do that:
const countBs = (str, i = 0) =>
i >= str.length
? 0
: countBs(str, i + 1) + (str[i].toUpperCase() === "B");
Or if you plan to use it for very very long strings, allow for TCO:
function countBs(str, i = 0, count = 0) {
if(i >= str.length) return count;
return countBs(str, i + 1, count + (str[i].toUpperCase === "B"));
}
answered Nov 14 '18 at 16:07
Jonas WilmsJonas Wilms
58.1k43051
1
Could it be donereturn cursor(++i, counted)
– George Jempty
Nov 14 '18 at 16:11
@george yes, but I don't see any sense in increasingi...
– Jonas Wilms
Nov 14 '18 at 16:12
@georg right, edited
– Jonas Wilms
Nov 14 '18 at 16:21
@JonasWilms So "++" operator just works locally? Is that why it works fine withindex++in for loop?
– Saeed Ahadian
Nov 14 '18 at 16:30
@saeed it evaluates toindex, and then afterwards increases it.
– Jonas Wilms
Nov 14 '18 at 16:41
add a comment |
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1 Answer
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votes
because
return cursor(i++, counted++);
has to be
return cursor(i + 1, counted + 1);
(as you want to increase the value passed recursively, not the local variable i)
How i would do that:
const countBs = (str, i = 0) =>
i >= str.length
? 0
: countBs(str, i + 1) + (str[i].toUpperCase() === "B");
Or if you plan to use it for very very long strings, allow for TCO:
function countBs(str, i = 0, count = 0) {
if(i >= str.length) return count;
return countBs(str, i + 1, count + (str[i].toUpperCase === "B"));
}
answered Nov 14 '18 at 16:07
Jonas WilmsJonas Wilms
58.1k43051
1
Could it be donereturn cursor(++i, counted)
– George Jempty
Nov 14 '18 at 16:11
@george yes, but I don't see any sense in increasingi...
– Jonas Wilms
Nov 14 '18 at 16:12
@georg right, edited
– Jonas Wilms
Nov 14 '18 at 16:21
@JonasWilms So "++" operator just works locally? Is that why it works fine withindex++in for loop?
– Saeed Ahadian
Nov 14 '18 at 16:30
@saeed it evaluates toindex, and then afterwards increases it.
– Jonas Wilms
Nov 14 '18 at 16:41
add a comment |
because
return cursor(i++, counted++);
has to be
return cursor(i + 1, counted + 1);
(as you want to increase the value passed recursively, not the local variable i)
How i would do that:
const countBs = (str, i = 0) =>
i >= str.length
? 0
: countBs(str, i + 1) + (str[i].toUpperCase() === "B");
Or if you plan to use it for very very long strings, allow for TCO:
function countBs(str, i = 0, count = 0) {
if(i >= str.length) return count;
return countBs(str, i + 1, count + (str[i].toUpperCase === "B"));
}
answered Nov 14 '18 at 16:07
Jonas WilmsJonas Wilms
58.1k43051
1
Could it be donereturn cursor(++i, counted)
– George Jempty
Nov 14 '18 at 16:11
@george yes, but I don't see any sense in increasingi...
– Jonas Wilms
Nov 14 '18 at 16:12
@georg right, edited
– Jonas Wilms
Nov 14 '18 at 16:21
@JonasWilms So "++" operator just works locally? Is that why it works fine withindex++in for loop?
– Saeed Ahadian
Nov 14 '18 at 16:30
@saeed it evaluates toindex, and then afterwards increases it.
– Jonas Wilms
Nov 14 '18 at 16:41
add a comment |
because
return cursor(i++, counted++);
has to be
return cursor(i + 1, counted + 1);
(as you want to increase the value passed recursively, not the local variable i)
How i would do that:
const countBs = (str, i = 0) =>
i >= str.length
? 0
: countBs(str, i + 1) + (str[i].toUpperCase() === "B");
Or if you plan to use it for very very long strings, allow for TCO:
function countBs(str, i = 0, count = 0) {
if(i >= str.length) return count;
return countBs(str, i + 1, count + (str[i].toUpperCase === "B"));
}
answered Nov 14 '18 at 16:07
Jonas WilmsJonas Wilms
58.1k43051
because
return cursor(i++, counted++);
has to be
return cursor(i + 1, counted + 1);
(as you want to increase the value passed recursively, not the local variable i)
How i would do that:
const countBs = (str, i = 0) =>
i >= str.length
? 0
: countBs(str, i + 1) + (str[i].toUpperCase() === "B");
Or if you plan to use it for very very long strings, allow for TCO:
function countBs(str, i = 0, count = 0) {
if(i >= str.length) return count;
return countBs(str, i + 1, count + (str[i].toUpperCase === "B"));
}
answered Nov 14 '18 at 16:07
Jonas WilmsJonas Wilms
58.1k43051
edited Nov 14 '18 at 16:21
answered Nov 14 '18 at 16:07
Jonas WilmsJonas Wilms
58.1k43051
answered Nov 14 '18 at 16:07
Jonas WilmsJonas Wilms
58.1k43051
answered Nov 14 '18 at 16:07
Jonas WilmsJonas Wilms
58.1k43051
58.1k43051
1
Could it be donereturn cursor(++i, counted)
– George Jempty
Nov 14 '18 at 16:11
@george yes, but I don't see any sense in increasingi...
– Jonas Wilms
Nov 14 '18 at 16:12
@georg right, edited
– Jonas Wilms
Nov 14 '18 at 16:21
@JonasWilms So "++" operator just works locally? Is that why it works fine withindex++in for loop?
– Saeed Ahadian
Nov 14 '18 at 16:30
@saeed it evaluates toindex, and then afterwards increases it.
– Jonas Wilms
Nov 14 '18 at 16:41
add a comment |
1
Could it be donereturn cursor(++i, counted)
– George Jempty
Nov 14 '18 at 16:11
@george yes, but I don't see any sense in increasingi...
– Jonas Wilms
Nov 14 '18 at 16:12
@georg right, edited
– Jonas Wilms
Nov 14 '18 at 16:21
@JonasWilms So "++" operator just works locally? Is that why it works fine withindex++in for loop?
– Saeed Ahadian
Nov 14 '18 at 16:30
@saeed it evaluates toindex, and then afterwards increases it.
– Jonas Wilms
Nov 14 '18 at 16:41
1
1
Could it be done
return cursor(++i, counted)– George Jempty
Nov 14 '18 at 16:11
Could it be done
return cursor(++i, counted)– George Jempty
Nov 14 '18 at 16:11
@george yes, but I don't see any sense in increasing
i ...– Jonas Wilms
Nov 14 '18 at 16:12
@george yes, but I don't see any sense in increasing
i ...– Jonas Wilms
Nov 14 '18 at 16:12
@georg right, edited
– Jonas Wilms
Nov 14 '18 at 16:21
@georg right, edited
– Jonas Wilms
Nov 14 '18 at 16:21
@JonasWilms So "++" operator just works locally? Is that why it works fine with
index++ in for loop?– Saeed Ahadian
Nov 14 '18 at 16:30
@JonasWilms So "++" operator just works locally? Is that why it works fine with
index++ in for loop?– Saeed Ahadian
Nov 14 '18 at 16:30
@saeed it evaluates to
index, and then afterwards increases it.– Jonas Wilms
Nov 14 '18 at 16:41
@saeed it evaluates to
index, and then afterwards increases it.– Jonas Wilms
Nov 14 '18 at 16:41
add a comment |
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1
Consider renaming
stringandcharto some not-so-system names.– Justinas
Nov 14 '18 at 16:08