My bash arithmetic always evaluates to 0, no matter what
I type in the script:
result=$(( ($1-32)*(5/9) ))
echo $result
and the result variable always evaluates to zero, no matter what value $1 is. How do I solve this problem?
bash
asked Jan 9 at 15:59
ThomasLMahoneyThomasLMahoney
72
add a comment |
I type in the script:
result=$(( ($1-32)*(5/9) ))
echo $result
and the result variable always evaluates to zero, no matter what value $1 is. How do I solve this problem?
bash
asked Jan 9 at 15:59
ThomasLMahoneyThomasLMahoney
72
add a comment |
I type in the script:
result=$(( ($1-32)*(5/9) ))
echo $result
and the result variable always evaluates to zero, no matter what value $1 is. How do I solve this problem?
bash
asked Jan 9 at 15:59
ThomasLMahoneyThomasLMahoney
72
I type in the script:
result=$(( ($1-32)*(5/9) ))
echo $result
and the result variable always evaluates to zero, no matter what value $1 is. How do I solve this problem?
bash
bash
asked Jan 9 at 15:59
ThomasLMahoneyThomasLMahoney
72
asked Jan 9 at 15:59
ThomasLMahoneyThomasLMahoney
72
asked Jan 9 at 15:59
ThomasLMahoneyThomasLMahoney
72
asked Jan 9 at 15:59
ThomasLMahoneyThomasLMahoney
72
asked Jan 9 at 15:59
ThomasLMahoneyThomasLMahoney
72
72
add a comment |
add a comment |
1 Answer
1
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The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.
Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas
$ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
0
if you allow * and / to have their natural precedence
$ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
1
because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.
$ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
1.65
answered Jan 9 at 16:37
steeldriversteeldriver
66.3k11106179
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.
Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas
$ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
0
if you allow * and / to have their natural precedence
$ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
1
because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.
$ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
1.65
answered Jan 9 at 16:37
steeldriversteeldriver
66.3k11106179
add a comment |
The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.
Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas
$ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
0
if you allow * and / to have their natural precedence
$ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
1
because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.
$ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
1.65
answered Jan 9 at 16:37
steeldriversteeldriver
66.3k11106179
add a comment |
The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.
Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas
$ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
0
if you allow * and / to have their natural precedence
$ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
1
because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.
$ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
1.65
answered Jan 9 at 16:37
steeldriversteeldriver
66.3k11106179
The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.
Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas
$ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
0
if you allow * and / to have their natural precedence
$ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
1
because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.
$ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
1.65
answered Jan 9 at 16:37
steeldriversteeldriver
66.3k11106179
answered Jan 9 at 16:37
steeldriversteeldriver
66.3k11106179
answered Jan 9 at 16:37
steeldriversteeldriver
66.3k11106179
answered Jan 9 at 16:37
steeldriversteeldriver
66.3k11106179
66.3k11106179
add a comment |
add a comment |
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