Weird 'string' declared here error. Trying to program pyramid of asterisks
I'm new to C++, so I'm still trying to figure out how to do basic debugging from errors printed on the terminal. I'm trying to print a pyramid of asterisks (*), but I keep getting this error that says "'string' declared here". /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/include/c++/v1/iosfwd:194:65: note:
'string' declared here
typedef basic_string, allocator > string;
^
1 error generated.
I've tried looking up how to read some of these errors in C++, but haven't been able to find a useful guide that isn't written in some alien technical gibberish. So if you could dumb down the explanation a little that would be great.
#include <iostream>
#include <vector>
#include <string>
#include <fstream>
using namespace std;
String printAst(int number){
for(int i = 0; i < number; i++){
cout << "* ";
}
}
int main(){
printAst(3);
//***
//**
//*
return 0;
}
I have the expected print out listed in //
c++
asked Jan 1 at 7:37
pancham2016pancham2016
11
|
show 2 more comments
I'm new to C++, so I'm still trying to figure out how to do basic debugging from errors printed on the terminal. I'm trying to print a pyramid of asterisks (*), but I keep getting this error that says "'string' declared here". /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/include/c++/v1/iosfwd:194:65: note:
'string' declared here
typedef basic_string, allocator > string;
^
1 error generated.
I've tried looking up how to read some of these errors in C++, but haven't been able to find a useful guide that isn't written in some alien technical gibberish. So if you could dumb down the explanation a little that would be great.
#include <iostream>
#include <vector>
#include <string>
#include <fstream>
using namespace std;
String printAst(int number){
for(int i = 0; i < number; i++){
cout << "* ";
}
}
int main(){
printAst(3);
//***
//**
//*
return 0;
}
I have the expected print out listed in //
c++
asked Jan 1 at 7:37
pancham2016pancham2016
11
One obvious potential problem is that theprintAst()function returns a string, but you have no return statement.
– Tim Biegeleisen
Jan 1 at 7:40
That isn't an error message it's a note about the error message on the line above in the compiler output
– Alan Birtles
Jan 1 at 7:41
1
The error message probably says something like "String is undefined, did you mean string?"
– Alan Birtles
Jan 1 at 7:43
You need to print 'n' to go to the next line.
– stark
Jan 1 at 14:37
@TimBiegeleisen I was confused about the return type, since I couldn't figure out whether cout counted as a string return, or whether I should have made the return type void
– pancham2016
Jan 2 at 23:57
|
show 2 more comments
I'm new to C++, so I'm still trying to figure out how to do basic debugging from errors printed on the terminal. I'm trying to print a pyramid of asterisks (*), but I keep getting this error that says "'string' declared here". /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/include/c++/v1/iosfwd:194:65: note:
'string' declared here
typedef basic_string, allocator > string;
^
1 error generated.
I've tried looking up how to read some of these errors in C++, but haven't been able to find a useful guide that isn't written in some alien technical gibberish. So if you could dumb down the explanation a little that would be great.
#include <iostream>
#include <vector>
#include <string>
#include <fstream>
using namespace std;
String printAst(int number){
for(int i = 0; i < number; i++){
cout << "* ";
}
}
int main(){
printAst(3);
//***
//**
//*
return 0;
}
I have the expected print out listed in //
c++
asked Jan 1 at 7:37
pancham2016pancham2016
11
I'm new to C++, so I'm still trying to figure out how to do basic debugging from errors printed on the terminal. I'm trying to print a pyramid of asterisks (*), but I keep getting this error that says "'string' declared here". /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/include/c++/v1/iosfwd:194:65: note:
'string' declared here
typedef basic_string, allocator > string;
^
1 error generated.
I've tried looking up how to read some of these errors in C++, but haven't been able to find a useful guide that isn't written in some alien technical gibberish. So if you could dumb down the explanation a little that would be great.
#include <iostream>
#include <vector>
#include <string>
#include <fstream>
using namespace std;
String printAst(int number){
for(int i = 0; i < number; i++){
cout << "* ";
}
}
int main(){
printAst(3);
//***
//**
//*
return 0;
}
I have the expected print out listed in //
c++
c++
asked Jan 1 at 7:37
pancham2016pancham2016
11
asked Jan 1 at 7:37
pancham2016pancham2016
11
asked Jan 1 at 7:37
pancham2016pancham2016
11
asked Jan 1 at 7:37
pancham2016pancham2016
11
asked Jan 1 at 7:37
pancham2016pancham2016
11
11
One obvious potential problem is that theprintAst()function returns a string, but you have no return statement.
– Tim Biegeleisen
Jan 1 at 7:40
That isn't an error message it's a note about the error message on the line above in the compiler output
– Alan Birtles
Jan 1 at 7:41
1
The error message probably says something like "String is undefined, did you mean string?"
– Alan Birtles
Jan 1 at 7:43
You need to print 'n' to go to the next line.
– stark
Jan 1 at 14:37
@TimBiegeleisen I was confused about the return type, since I couldn't figure out whether cout counted as a string return, or whether I should have made the return type void
– pancham2016
Jan 2 at 23:57
|
show 2 more comments
One obvious potential problem is that theprintAst()function returns a string, but you have no return statement.
– Tim Biegeleisen
Jan 1 at 7:40
That isn't an error message it's a note about the error message on the line above in the compiler output
– Alan Birtles
Jan 1 at 7:41
1
The error message probably says something like "String is undefined, did you mean string?"
– Alan Birtles
Jan 1 at 7:43
You need to print 'n' to go to the next line.
– stark
Jan 1 at 14:37
@TimBiegeleisen I was confused about the return type, since I couldn't figure out whether cout counted as a string return, or whether I should have made the return type void
– pancham2016
Jan 2 at 23:57
One obvious potential problem is that the
printAst() function returns a string, but you have no return statement.– Tim Biegeleisen
Jan 1 at 7:40
One obvious potential problem is that the
printAst() function returns a string, but you have no return statement.– Tim Biegeleisen
Jan 1 at 7:40
That isn't an error message it's a note about the error message on the line above in the compiler output
– Alan Birtles
Jan 1 at 7:41
That isn't an error message it's a note about the error message on the line above in the compiler output
– Alan Birtles
Jan 1 at 7:41
1
1
The error message probably says something like "String is undefined, did you mean string?"
– Alan Birtles
Jan 1 at 7:43
The error message probably says something like "String is undefined, did you mean string?"
– Alan Birtles
Jan 1 at 7:43
You need to print 'n' to go to the next line.
– stark
Jan 1 at 14:37
You need to print 'n' to go to the next line.
– stark
Jan 1 at 14:37
@TimBiegeleisen I was confused about the return type, since I couldn't figure out whether cout counted as a string return, or whether I should have made the return type void
– pancham2016
Jan 2 at 23:57
@TimBiegeleisen I was confused about the return type, since I couldn't figure out whether cout counted as a string return, or whether I should have made the return type void
– pancham2016
Jan 2 at 23:57
|
show 2 more comments
1 Answer
1
active
oldest
votes
Here is a corrected version of your code. The immediate cause of your current errors appears to have been the String return type of the printAst() function. If you wanted to return a string, then you probably intended to use std::string, not String. But, even after fixing this, I noticed that your logic for printing the pyramid also had a problem, so I fixed that too.
void printAst (int number) {
for (int i=0; i < number; ++i) {
for (int j=0; j < number-i; ++j) {
cout << "* ";
}
cout << "n";
}
}
int main() {
printAst(3);
return 0;
}
* * *
* *
*
The major changes I made include using a double for loop to print the inverse pyramid. The logic is that there are number levels to the pyramid, and at each level we print number count of asterisks.
Also, since printAst does not return anything, I changed the return type to void.
answered Jan 1 at 7:43
Tim BiegeleisenTim Biegeleisen
227k1394147
Ya, I ended up reaching the same conclusion regarding the typing being void and using the nested for loop. What I'm still confused about is how I know the type should be void. I thought initially since I'm printing a string "*" the return type should be string, but now I'm not sure.
– pancham2016
Jan 3 at 0:04
TheprintAstfunction does not actually return any value, thus it should be void. That is prints out a string has nothing to do with its return behavior.
– Tim Biegeleisen
Jan 3 at 0:12
add a comment |
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1 Answer
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Here is a corrected version of your code. The immediate cause of your current errors appears to have been the String return type of the printAst() function. If you wanted to return a string, then you probably intended to use std::string, not String. But, even after fixing this, I noticed that your logic for printing the pyramid also had a problem, so I fixed that too.
void printAst (int number) {
for (int i=0; i < number; ++i) {
for (int j=0; j < number-i; ++j) {
cout << "* ";
}
cout << "n";
}
}
int main() {
printAst(3);
return 0;
}
* * *
* *
*
The major changes I made include using a double for loop to print the inverse pyramid. The logic is that there are number levels to the pyramid, and at each level we print number count of asterisks.
Also, since printAst does not return anything, I changed the return type to void.
answered Jan 1 at 7:43
Tim BiegeleisenTim Biegeleisen
227k1394147
Ya, I ended up reaching the same conclusion regarding the typing being void and using the nested for loop. What I'm still confused about is how I know the type should be void. I thought initially since I'm printing a string "*" the return type should be string, but now I'm not sure.
– pancham2016
Jan 3 at 0:04
TheprintAstfunction does not actually return any value, thus it should be void. That is prints out a string has nothing to do with its return behavior.
– Tim Biegeleisen
Jan 3 at 0:12
add a comment |
Here is a corrected version of your code. The immediate cause of your current errors appears to have been the String return type of the printAst() function. If you wanted to return a string, then you probably intended to use std::string, not String. But, even after fixing this, I noticed that your logic for printing the pyramid also had a problem, so I fixed that too.
void printAst (int number) {
for (int i=0; i < number; ++i) {
for (int j=0; j < number-i; ++j) {
cout << "* ";
}
cout << "n";
}
}
int main() {
printAst(3);
return 0;
}
* * *
* *
*
The major changes I made include using a double for loop to print the inverse pyramid. The logic is that there are number levels to the pyramid, and at each level we print number count of asterisks.
Also, since printAst does not return anything, I changed the return type to void.
answered Jan 1 at 7:43
Tim BiegeleisenTim Biegeleisen
227k1394147
Ya, I ended up reaching the same conclusion regarding the typing being void and using the nested for loop. What I'm still confused about is how I know the type should be void. I thought initially since I'm printing a string "*" the return type should be string, but now I'm not sure.
– pancham2016
Jan 3 at 0:04
TheprintAstfunction does not actually return any value, thus it should be void. That is prints out a string has nothing to do with its return behavior.
– Tim Biegeleisen
Jan 3 at 0:12
add a comment |
Here is a corrected version of your code. The immediate cause of your current errors appears to have been the String return type of the printAst() function. If you wanted to return a string, then you probably intended to use std::string, not String. But, even after fixing this, I noticed that your logic for printing the pyramid also had a problem, so I fixed that too.
void printAst (int number) {
for (int i=0; i < number; ++i) {
for (int j=0; j < number-i; ++j) {
cout << "* ";
}
cout << "n";
}
}
int main() {
printAst(3);
return 0;
}
* * *
* *
*
The major changes I made include using a double for loop to print the inverse pyramid. The logic is that there are number levels to the pyramid, and at each level we print number count of asterisks.
Also, since printAst does not return anything, I changed the return type to void.
answered Jan 1 at 7:43
Tim BiegeleisenTim Biegeleisen
227k1394147
Here is a corrected version of your code. The immediate cause of your current errors appears to have been the String return type of the printAst() function. If you wanted to return a string, then you probably intended to use std::string, not String. But, even after fixing this, I noticed that your logic for printing the pyramid also had a problem, so I fixed that too.
void printAst (int number) {
for (int i=0; i < number; ++i) {
for (int j=0; j < number-i; ++j) {
cout << "* ";
}
cout << "n";
}
}
int main() {
printAst(3);
return 0;
}
* * *
* *
*
The major changes I made include using a double for loop to print the inverse pyramid. The logic is that there are number levels to the pyramid, and at each level we print number count of asterisks.
Also, since printAst does not return anything, I changed the return type to void.
answered Jan 1 at 7:43
Tim BiegeleisenTim Biegeleisen
227k1394147
answered Jan 1 at 7:43
Tim BiegeleisenTim Biegeleisen
227k1394147
answered Jan 1 at 7:43
Tim BiegeleisenTim Biegeleisen
227k1394147
answered Jan 1 at 7:43
Tim BiegeleisenTim Biegeleisen
227k1394147
227k1394147
Ya, I ended up reaching the same conclusion regarding the typing being void and using the nested for loop. What I'm still confused about is how I know the type should be void. I thought initially since I'm printing a string "*" the return type should be string, but now I'm not sure.
– pancham2016
Jan 3 at 0:04
TheprintAstfunction does not actually return any value, thus it should be void. That is prints out a string has nothing to do with its return behavior.
– Tim Biegeleisen
Jan 3 at 0:12
add a comment |
Ya, I ended up reaching the same conclusion regarding the typing being void and using the nested for loop. What I'm still confused about is how I know the type should be void. I thought initially since I'm printing a string "*" the return type should be string, but now I'm not sure.
– pancham2016
Jan 3 at 0:04
TheprintAstfunction does not actually return any value, thus it should be void. That is prints out a string has nothing to do with its return behavior.
– Tim Biegeleisen
Jan 3 at 0:12
Ya, I ended up reaching the same conclusion regarding the typing being void and using the nested for loop. What I'm still confused about is how I know the type should be void. I thought initially since I'm printing a string "*" the return type should be string, but now I'm not sure.
– pancham2016
Jan 3 at 0:04
Ya, I ended up reaching the same conclusion regarding the typing being void and using the nested for loop. What I'm still confused about is how I know the type should be void. I thought initially since I'm printing a string "*" the return type should be string, but now I'm not sure.
– pancham2016
Jan 3 at 0:04
The
printAst function does not actually return any value, thus it should be void. That is prints out a string has nothing to do with its return behavior.– Tim Biegeleisen
Jan 3 at 0:12
The
printAst function does not actually return any value, thus it should be void. That is prints out a string has nothing to do with its return behavior.– Tim Biegeleisen
Jan 3 at 0:12
add a comment |
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One obvious potential problem is that the
printAst()function returns a string, but you have no return statement.– Tim Biegeleisen
Jan 1 at 7:40
That isn't an error message it's a note about the error message on the line above in the compiler output
– Alan Birtles
Jan 1 at 7:41
1
The error message probably says something like "String is undefined, did you mean string?"
– Alan Birtles
Jan 1 at 7:43
You need to print 'n' to go to the next line.
– stark
Jan 1 at 14:37
@TimBiegeleisen I was confused about the return type, since I couldn't figure out whether cout counted as a string return, or whether I should have made the return type void
– pancham2016
Jan 2 at 23:57