Overflowed value prints non-overflowed value on Arduino
First of all, I know the problem with the code and how to get it to work. I'm mainly looking for an explanation why my output is what it is.
The following piece of code replicates the behaviour:
void setup() {
Serial.begin(9600);
}
void loop() {
Serial.println("start");
for(int i = 0; i < 70000; i++) {
if((i % 2000) == 0)
Serial.println(i);
}
}
Obviously the for loop will run forever because i will overflow at 32,767. I would expect it to overflow and print -32000.
Expected| Actually printed
0 | 0
2000 | 2000
4000 | 4000
... | ...
30000 | 30000
32000 | 32000
-32000 | 33536
-30000 | 35536
It looks like it prints the actual iterations, since if you overflow and count to -32000 you would have 33536 iterations, but I can't figure out how it's able to print the 33536.
The same thing happens every 65536 iterations:
95536 | 161072 | 226608 | 292144 | 357680
97536 | 163072 | 228608 | 294144 | 359680
99072 | 164608 | 230144 | 295680 | 361216
101072 | 166608 | 232144 | 297680 | 363216
EDIT
When I change the loop to add 10.000 every iteration and only print every 1.000.000 to speed it up the Arduino crashes (or at least, the prints stop) at '2.147.000.000'. Which seems to point to the 32-bit idea of svtag**
EDIT2
The edits I made for the 2.147.000.000-check:
void loop() {
Serial.println("start");
for(int i = 0; i < 70000; i+=10000) {
if((i % 1000000) == 0)
Serial.println(i);
}
}
They work in the same trend with the previous example, printing 0, 1000000, 2000000,...
However, when I update the AVR package from 1.6.17 to the latest 1.6.23 I get only 0's. The original example (% 2000 & i++) still gives the same output.
c++ arduino arduino-uno
edited Nov 15 '18 at 18:27
gre_gor
4,13892631
asked Nov 15 '18 at 15:08
LonefishLonefish
3271620
|
show 11 more comments
First of all, I know the problem with the code and how to get it to work. I'm mainly looking for an explanation why my output is what it is.
The following piece of code replicates the behaviour:
void setup() {
Serial.begin(9600);
}
void loop() {
Serial.println("start");
for(int i = 0; i < 70000; i++) {
if((i % 2000) == 0)
Serial.println(i);
}
}
Obviously the for loop will run forever because i will overflow at 32,767. I would expect it to overflow and print -32000.
Expected| Actually printed
0 | 0
2000 | 2000
4000 | 4000
... | ...
30000 | 30000
32000 | 32000
-32000 | 33536
-30000 | 35536
It looks like it prints the actual iterations, since if you overflow and count to -32000 you would have 33536 iterations, but I can't figure out how it's able to print the 33536.
The same thing happens every 65536 iterations:
95536 | 161072 | 226608 | 292144 | 357680
97536 | 163072 | 228608 | 294144 | 359680
99072 | 164608 | 230144 | 295680 | 361216
101072 | 166608 | 232144 | 297680 | 363216
EDIT
When I change the loop to add 10.000 every iteration and only print every 1.000.000 to speed it up the Arduino crashes (or at least, the prints stop) at '2.147.000.000'. Which seems to point to the 32-bit idea of svtag**
EDIT2
The edits I made for the 2.147.000.000-check:
void loop() {
Serial.println("start");
for(int i = 0; i < 70000; i+=10000) {
if((i % 1000000) == 0)
Serial.println(i);
}
}
They work in the same trend with the previous example, printing 0, 1000000, 2000000,...
However, when I update the AVR package from 1.6.17 to the latest 1.6.23 I get only 0's. The original example (% 2000 & i++) still gives the same output.
c++ arduino arduino-uno
edited Nov 15 '18 at 18:27
gre_gor
4,13892631
asked Nov 15 '18 at 15:08
LonefishLonefish
3271620
This can be because of the difference inintdefinition on the Arduino and PC. Serially Arduino will send the binary equivalent of (int)33536 in 16bit, but if your PC interpret it as 32bit, then you should get the actual number. Just a hypothesis, I will check it as soon as i get the hardware.
– svtag
Nov 15 '18 at 15:21
I've also been thinking something along those lines, but then it shouldn't be able to print numbers larger than 65535. Because in 16 bit that would be the max value the arduino could output. Somehow it has some sort of shadowcopy that is 32bit it seems..
– Lonefish
Nov 15 '18 at 15:50
1
what Arduino board? you are on some 32bit board
– Juraj
Nov 15 '18 at 16:05
1
@svtag, please. it would be number of digits. the conversion to text is done on MCU
– Juraj
Nov 15 '18 at 16:22
1
written to Serial. in case of number, after conversion to text. number of digits. +2 for rn
– Juraj
Nov 15 '18 at 16:27
|
show 11 more comments
First of all, I know the problem with the code and how to get it to work. I'm mainly looking for an explanation why my output is what it is.
The following piece of code replicates the behaviour:
void setup() {
Serial.begin(9600);
}
void loop() {
Serial.println("start");
for(int i = 0; i < 70000; i++) {
if((i % 2000) == 0)
Serial.println(i);
}
}
Obviously the for loop will run forever because i will overflow at 32,767. I would expect it to overflow and print -32000.
Expected| Actually printed
0 | 0
2000 | 2000
4000 | 4000
... | ...
30000 | 30000
32000 | 32000
-32000 | 33536
-30000 | 35536
It looks like it prints the actual iterations, since if you overflow and count to -32000 you would have 33536 iterations, but I can't figure out how it's able to print the 33536.
The same thing happens every 65536 iterations:
95536 | 161072 | 226608 | 292144 | 357680
97536 | 163072 | 228608 | 294144 | 359680
99072 | 164608 | 230144 | 295680 | 361216
101072 | 166608 | 232144 | 297680 | 363216
EDIT
When I change the loop to add 10.000 every iteration and only print every 1.000.000 to speed it up the Arduino crashes (or at least, the prints stop) at '2.147.000.000'. Which seems to point to the 32-bit idea of svtag**
EDIT2
The edits I made for the 2.147.000.000-check:
void loop() {
Serial.println("start");
for(int i = 0; i < 70000; i+=10000) {
if((i % 1000000) == 0)
Serial.println(i);
}
}
They work in the same trend with the previous example, printing 0, 1000000, 2000000,...
However, when I update the AVR package from 1.6.17 to the latest 1.6.23 I get only 0's. The original example (% 2000 & i++) still gives the same output.
c++ arduino arduino-uno
edited Nov 15 '18 at 18:27
gre_gor
4,13892631
asked Nov 15 '18 at 15:08
LonefishLonefish
3271620
First of all, I know the problem with the code and how to get it to work. I'm mainly looking for an explanation why my output is what it is.
The following piece of code replicates the behaviour:
void setup() {
Serial.begin(9600);
}
void loop() {
Serial.println("start");
for(int i = 0; i < 70000; i++) {
if((i % 2000) == 0)
Serial.println(i);
}
}
Obviously the for loop will run forever because i will overflow at 32,767. I would expect it to overflow and print -32000.
Expected| Actually printed
0 | 0
2000 | 2000
4000 | 4000
... | ...
30000 | 30000
32000 | 32000
-32000 | 33536
-30000 | 35536
It looks like it prints the actual iterations, since if you overflow and count to -32000 you would have 33536 iterations, but I can't figure out how it's able to print the 33536.
The same thing happens every 65536 iterations:
95536 | 161072 | 226608 | 292144 | 357680
97536 | 163072 | 228608 | 294144 | 359680
99072 | 164608 | 230144 | 295680 | 361216
101072 | 166608 | 232144 | 297680 | 363216
EDIT
When I change the loop to add 10.000 every iteration and only print every 1.000.000 to speed it up the Arduino crashes (or at least, the prints stop) at '2.147.000.000'. Which seems to point to the 32-bit idea of svtag**
EDIT2
The edits I made for the 2.147.000.000-check:
void loop() {
Serial.println("start");
for(int i = 0; i < 70000; i+=10000) {
if((i % 1000000) == 0)
Serial.println(i);
}
}
They work in the same trend with the previous example, printing 0, 1000000, 2000000,...
However, when I update the AVR package from 1.6.17 to the latest 1.6.23 I get only 0's. The original example (% 2000 & i++) still gives the same output.
c++ arduino arduino-uno
c++ arduino arduino-uno
edited Nov 15 '18 at 18:27
gre_gor
4,13892631
asked Nov 15 '18 at 15:08
LonefishLonefish
3271620
edited Nov 15 '18 at 18:27
gre_gor
4,13892631
asked Nov 15 '18 at 15:08
LonefishLonefish
3271620
edited Nov 15 '18 at 18:27
gre_gor
4,13892631
edited Nov 15 '18 at 18:27
gre_gor
4,13892631
edited Nov 15 '18 at 18:27
gre_gor
4,13892631
4,13892631
asked Nov 15 '18 at 15:08
LonefishLonefish
3271620
asked Nov 15 '18 at 15:08
LonefishLonefish
3271620
asked Nov 15 '18 at 15:08
LonefishLonefish
3271620
3271620
This can be because of the difference inintdefinition on the Arduino and PC. Serially Arduino will send the binary equivalent of (int)33536 in 16bit, but if your PC interpret it as 32bit, then you should get the actual number. Just a hypothesis, I will check it as soon as i get the hardware.
– svtag
Nov 15 '18 at 15:21
I've also been thinking something along those lines, but then it shouldn't be able to print numbers larger than 65535. Because in 16 bit that would be the max value the arduino could output. Somehow it has some sort of shadowcopy that is 32bit it seems..
– Lonefish
Nov 15 '18 at 15:50
1
what Arduino board? you are on some 32bit board
– Juraj
Nov 15 '18 at 16:05
1
@svtag, please. it would be number of digits. the conversion to text is done on MCU
– Juraj
Nov 15 '18 at 16:22
1
written to Serial. in case of number, after conversion to text. number of digits. +2 for rn
– Juraj
Nov 15 '18 at 16:27
|
show 11 more comments
This can be because of the difference inintdefinition on the Arduino and PC. Serially Arduino will send the binary equivalent of (int)33536 in 16bit, but if your PC interpret it as 32bit, then you should get the actual number. Just a hypothesis, I will check it as soon as i get the hardware.
– svtag
Nov 15 '18 at 15:21
I've also been thinking something along those lines, but then it shouldn't be able to print numbers larger than 65535. Because in 16 bit that would be the max value the arduino could output. Somehow it has some sort of shadowcopy that is 32bit it seems..
– Lonefish
Nov 15 '18 at 15:50
1
what Arduino board? you are on some 32bit board
– Juraj
Nov 15 '18 at 16:05
1
@svtag, please. it would be number of digits. the conversion to text is done on MCU
– Juraj
Nov 15 '18 at 16:22
1
written to Serial. in case of number, after conversion to text. number of digits. +2 for rn
– Juraj
Nov 15 '18 at 16:27
This can be because of the difference in
int definition on the Arduino and PC. Serially Arduino will send the binary equivalent of (int)33536 in 16bit, but if your PC interpret it as 32bit, then you should get the actual number. Just a hypothesis, I will check it as soon as i get the hardware.– svtag
Nov 15 '18 at 15:21
This can be because of the difference in
int definition on the Arduino and PC. Serially Arduino will send the binary equivalent of (int)33536 in 16bit, but if your PC interpret it as 32bit, then you should get the actual number. Just a hypothesis, I will check it as soon as i get the hardware.– svtag
Nov 15 '18 at 15:21
I've also been thinking something along those lines, but then it shouldn't be able to print numbers larger than 65535. Because in 16 bit that would be the max value the arduino could output. Somehow it has some sort of shadowcopy that is 32bit it seems..
– Lonefish
Nov 15 '18 at 15:50
I've also been thinking something along those lines, but then it shouldn't be able to print numbers larger than 65535. Because in 16 bit that would be the max value the arduino could output. Somehow it has some sort of shadowcopy that is 32bit it seems..
– Lonefish
Nov 15 '18 at 15:50
1
1
what Arduino board? you are on some 32bit board
– Juraj
Nov 15 '18 at 16:05
what Arduino board? you are on some 32bit board
– Juraj
Nov 15 '18 at 16:05
1
1
@svtag, please. it would be number of digits. the conversion to text is done on MCU
– Juraj
Nov 15 '18 at 16:22
@svtag, please. it would be number of digits. the conversion to text is done on MCU
– Juraj
Nov 15 '18 at 16:22
1
1
written to Serial. in case of number, after conversion to text. number of digits. +2 for rn
– Juraj
Nov 15 '18 at 16:27
written to Serial. in case of number, after conversion to text. number of digits. +2 for rn
– Juraj
Nov 15 '18 at 16:27
|
show 11 more comments
1 Answer
1
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oldest
votes
The compiler may have automatically casted the i to usigned int when you are doing the println or the value is going to the wrong overload. Try using Serial.println((int)i);
answered Nov 15 '18 at 16:31
Branislav BrzakBranislav Brzak
93
It manages to count beyond 65535, it counts until 2.147.000.000 (printing only millions), so it's a 32bit value somehow..
– Lonefish
Nov 15 '18 at 16:33
add a comment |
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The compiler may have automatically casted the i to usigned int when you are doing the println or the value is going to the wrong overload. Try using Serial.println((int)i);
answered Nov 15 '18 at 16:31
Branislav BrzakBranislav Brzak
93
It manages to count beyond 65535, it counts until 2.147.000.000 (printing only millions), so it's a 32bit value somehow..
– Lonefish
Nov 15 '18 at 16:33
add a comment |
The compiler may have automatically casted the i to usigned int when you are doing the println or the value is going to the wrong overload. Try using Serial.println((int)i);
answered Nov 15 '18 at 16:31
Branislav BrzakBranislav Brzak
93
It manages to count beyond 65535, it counts until 2.147.000.000 (printing only millions), so it's a 32bit value somehow..
– Lonefish
Nov 15 '18 at 16:33
add a comment |
The compiler may have automatically casted the i to usigned int when you are doing the println or the value is going to the wrong overload. Try using Serial.println((int)i);
answered Nov 15 '18 at 16:31
Branislav BrzakBranislav Brzak
93
The compiler may have automatically casted the i to usigned int when you are doing the println or the value is going to the wrong overload. Try using Serial.println((int)i);
answered Nov 15 '18 at 16:31
Branislav BrzakBranislav Brzak
93
answered Nov 15 '18 at 16:31
Branislav BrzakBranislav Brzak
93
answered Nov 15 '18 at 16:31
Branislav BrzakBranislav Brzak
93
answered Nov 15 '18 at 16:31
Branislav BrzakBranislav Brzak
93
93
It manages to count beyond 65535, it counts until 2.147.000.000 (printing only millions), so it's a 32bit value somehow..
– Lonefish
Nov 15 '18 at 16:33
add a comment |
It manages to count beyond 65535, it counts until 2.147.000.000 (printing only millions), so it's a 32bit value somehow..
– Lonefish
Nov 15 '18 at 16:33
It manages to count beyond 65535, it counts until 2.147.000.000 (printing only millions), so it's a 32bit value somehow..
– Lonefish
Nov 15 '18 at 16:33
It manages to count beyond 65535, it counts until 2.147.000.000 (printing only millions), so it's a 32bit value somehow..
– Lonefish
Nov 15 '18 at 16:33
add a comment |
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This can be because of the difference in
intdefinition on the Arduino and PC. Serially Arduino will send the binary equivalent of (int)33536 in 16bit, but if your PC interpret it as 32bit, then you should get the actual number. Just a hypothesis, I will check it as soon as i get the hardware.– svtag
Nov 15 '18 at 15:21
I've also been thinking something along those lines, but then it shouldn't be able to print numbers larger than 65535. Because in 16 bit that would be the max value the arduino could output. Somehow it has some sort of shadowcopy that is 32bit it seems..
– Lonefish
Nov 15 '18 at 15:50
1
what Arduino board? you are on some 32bit board
– Juraj
Nov 15 '18 at 16:05
1
@svtag, please. it would be number of digits. the conversion to text is done on MCU
– Juraj
Nov 15 '18 at 16:22
1
written to Serial. in case of number, after conversion to text. number of digits. +2 for rn
– Juraj
Nov 15 '18 at 16:27