Arduino: incorrect calculation of long integer
I'm doing a simple calculation with integers (on Arduino with ESP8266 12E), but I can't get the expected result and can't find the error. Can someone guide me?
#define A 200
#define B A * 62
#define C 500
void setup() {
Serial.begin(9600);
Serial.println("");
unsigned long aux = 0;
aux = (B * 500) / C; // (12400 * 500) / 500 = 12400
Serial.printf("aux = %dn", aux);
aux = aux * C; // 12400 * 500 = 6200000
Serial.printf("aux = %dn", aux);
// ERROR: Should result in "500", but is resulting in "1922000"
aux = aux / B; // 6200000 / 12400 = 500
Serial.printf("aux = %dn", aux); // It's printing "1922000"
}
arduino esp8266
add a comment |
I'm doing a simple calculation with integers (on Arduino with ESP8266 12E), but I can't get the expected result and can't find the error. Can someone guide me?
#define A 200
#define B A * 62
#define C 500
void setup() {
Serial.begin(9600);
Serial.println("");
unsigned long aux = 0;
aux = (B * 500) / C; // (12400 * 500) / 500 = 12400
Serial.printf("aux = %dn", aux);
aux = aux * C; // 12400 * 500 = 6200000
Serial.printf("aux = %dn", aux);
// ERROR: Should result in "500", but is resulting in "1922000"
aux = aux / B; // 6200000 / 12400 = 500
Serial.printf("aux = %dn", aux); // It's printing "1922000"
}
arduino esp8266
add a comment |
I'm doing a simple calculation with integers (on Arduino with ESP8266 12E), but I can't get the expected result and can't find the error. Can someone guide me?
#define A 200
#define B A * 62
#define C 500
void setup() {
Serial.begin(9600);
Serial.println("");
unsigned long aux = 0;
aux = (B * 500) / C; // (12400 * 500) / 500 = 12400
Serial.printf("aux = %dn", aux);
aux = aux * C; // 12400 * 500 = 6200000
Serial.printf("aux = %dn", aux);
// ERROR: Should result in "500", but is resulting in "1922000"
aux = aux / B; // 6200000 / 12400 = 500
Serial.printf("aux = %dn", aux); // It's printing "1922000"
}
arduino esp8266
I'm doing a simple calculation with integers (on Arduino with ESP8266 12E), but I can't get the expected result and can't find the error. Can someone guide me?
#define A 200
#define B A * 62
#define C 500
void setup() {
Serial.begin(9600);
Serial.println("");
unsigned long aux = 0;
aux = (B * 500) / C; // (12400 * 500) / 500 = 12400
Serial.printf("aux = %dn", aux);
aux = aux * C; // 12400 * 500 = 6200000
Serial.printf("aux = %dn", aux);
// ERROR: Should result in "500", but is resulting in "1922000"
aux = aux / B; // 6200000 / 12400 = 500
Serial.printf("aux = %dn", aux); // It's printing "1922000"
}
arduino esp8266
arduino esp8266
asked 2 hours ago
wBB
1134
1134
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
1 hour ago
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
1 hour ago
add a comment |
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L
suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int
). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
1
...and the format specifier for anunsigned long
is%lu
not%d
- the compiler for the asker's esp8266 uses a 32-bitint
so they get away with some things they would not on an ATmega-based Arduino where anint
is the minimum 16 bit size allowed by the specification.
– Chris Stratton
8 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
1 hour ago
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
1 hour ago
add a comment |
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
1 hour ago
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
1 hour ago
add a comment |
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
In your #define of B you missed parenthesis (). Change your definition to:
#define B (A * 62)
Without parenthesis you first divide 6200000 by 200 and then multiply result by 62, which is not what you intend.
answered 1 hour ago
dmz
1365
1365
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
1 hour ago
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
1 hour ago
add a comment |
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
1 hour ago
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
1 hour ago
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
1 hour ago
Dude, you're 100% right. I've spent several hours trying to figure out what was wrong ... Thank you so much!
– wBB
1 hour ago
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
1 hour ago
@wBB remember that in C, macros are replaced in the code, exactly as you wrote them, in the preprocessor step before the code gets compiled. So it helps as a sanity-check in these cases to expand the macros yourself in your code to see if you're getting what you intended.
– brhans
1 hour ago
add a comment |
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L
suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int
). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
1
...and the format specifier for anunsigned long
is%lu
not%d
- the compiler for the asker's esp8266 uses a 32-bitint
so they get away with some things they would not on an ATmega-based Arduino where anint
is the minimum 16 bit size allowed by the specification.
– Chris Stratton
8 mins ago
add a comment |
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L
suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int
). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
1
...and the format specifier for anunsigned long
is%lu
not%d
- the compiler for the asker's esp8266 uses a 32-bitint
so they get away with some things they would not on an ATmega-based Arduino where anint
is the minimum 16 bit size allowed by the specification.
– Chris Stratton
8 mins ago
add a comment |
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L
suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int
). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
Fully-parenthesizing macros (as noted in answer by dmz) solves one class of problem.
Another thing you should do is, in any arithmetic expression which involves literal constants, use the L
suffix on at least one of the constants involved if there's any chance the result will exceed 32767 (the maximum guaranteed-representable value for int
). The type of an arithmetic operation in C is based on the types of the operands of that operation only; the type of the variable to which the result is assigned is irrelevant.
edited 3 mins ago
answered 11 mins ago
mlp
17215
17215
1
...and the format specifier for anunsigned long
is%lu
not%d
- the compiler for the asker's esp8266 uses a 32-bitint
so they get away with some things they would not on an ATmega-based Arduino where anint
is the minimum 16 bit size allowed by the specification.
– Chris Stratton
8 mins ago
add a comment |
1
...and the format specifier for anunsigned long
is%lu
not%d
- the compiler for the asker's esp8266 uses a 32-bitint
so they get away with some things they would not on an ATmega-based Arduino where anint
is the minimum 16 bit size allowed by the specification.
– Chris Stratton
8 mins ago
1
1
...and the format specifier for an
unsigned long
is %lu
not %d
- the compiler for the asker's esp8266 uses a 32-bit int
so they get away with some things they would not on an ATmega-based Arduino where an int
is the minimum 16 bit size allowed by the specification.– Chris Stratton
8 mins ago
...and the format specifier for an
unsigned long
is %lu
not %d
- the compiler for the asker's esp8266 uses a 32-bit int
so they get away with some things they would not on an ATmega-based Arduino where an int
is the minimum 16 bit size allowed by the specification.– Chris Stratton
8 mins ago
add a comment |
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