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Variable within a function does not accept its new value!


Why is my variable local in one 'while read' loop, but not in another seemingly similar loop?creating variable using variable value as part of new variable nameTest variable if its string or notValue assigned inside a function variable is always emptyAssign variable identifier in loop and echo its valueBASH Variable checker functionDifference between set variable “value” and set variable value?How to print the variable name along with its value?Declaring a variable that does not have inputSetting variable output from timeout






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}







0















I wrote the following function to sequentially add sum calculation into a variable.



RMS_Cal ()
{
num=`cat synt_tdiffs.dat | wc -l`
Sum=0
cat synt_tdiffs.dat | while read info; do
sta=`echo $info | cut -f 2 -d " "`
tds=`echo $info | cut -f 3 -d " "`
tdr=`grep -w "$sta" real_tdiffs.dat | cut -f 3 -d " "`
sum=`echo "($tds - $tdr) ^ 2" | bc -l`
Sum=`echo "$sum + $Sum" | bc -l`
done
echo $Sum
}


The problem is the "Sum" variable does not accept its new value after the while loop ends.



 RMS_Cal
++ cat synt_tdiffs.dat
++ wc -l
+ num=1
+ local Sum=0
+ cat synt_tdiffs.dat
+ read info
++ echo AHWZ ILBA 25.631
++ cut -f 2 -d ' '
+ sta=ILBA
++ echo AHWZ ILBA 25.631
++ cut -f 3 -d ' '
+ tds=25.631
++ grep -w ILBA real_tdiffs.dat
++ cut -f 3 -d ' '
+ tdr=527
++ echo '(25.631 - 527) ^ 2'
++ bc -l
+ sum=251370.874161
++ echo '251370.874161 + 0'
++ bc -l
+ Sum=251370.874161
+ read info
+ echo 0
0


How can I handle this problem?









share

























  • Possible duplicate of Why is my variable local in one 'while read' loop, but not in another seemingly similar loop?

    – muru
    3 mins ago


















0















I wrote the following function to sequentially add sum calculation into a variable.



RMS_Cal ()
{
num=`cat synt_tdiffs.dat | wc -l`
Sum=0
cat synt_tdiffs.dat | while read info; do
sta=`echo $info | cut -f 2 -d " "`
tds=`echo $info | cut -f 3 -d " "`
tdr=`grep -w "$sta" real_tdiffs.dat | cut -f 3 -d " "`
sum=`echo "($tds - $tdr) ^ 2" | bc -l`
Sum=`echo "$sum + $Sum" | bc -l`
done
echo $Sum
}


The problem is the "Sum" variable does not accept its new value after the while loop ends.



 RMS_Cal
++ cat synt_tdiffs.dat
++ wc -l
+ num=1
+ local Sum=0
+ cat synt_tdiffs.dat
+ read info
++ echo AHWZ ILBA 25.631
++ cut -f 2 -d ' '
+ sta=ILBA
++ echo AHWZ ILBA 25.631
++ cut -f 3 -d ' '
+ tds=25.631
++ grep -w ILBA real_tdiffs.dat
++ cut -f 3 -d ' '
+ tdr=527
++ echo '(25.631 - 527) ^ 2'
++ bc -l
+ sum=251370.874161
++ echo '251370.874161 + 0'
++ bc -l
+ Sum=251370.874161
+ read info
+ echo 0
0


How can I handle this problem?









share

























  • Possible duplicate of Why is my variable local in one 'while read' loop, but not in another seemingly similar loop?

    – muru
    3 mins ago














0












0








0








I wrote the following function to sequentially add sum calculation into a variable.



RMS_Cal ()
{
num=`cat synt_tdiffs.dat | wc -l`
Sum=0
cat synt_tdiffs.dat | while read info; do
sta=`echo $info | cut -f 2 -d " "`
tds=`echo $info | cut -f 3 -d " "`
tdr=`grep -w "$sta" real_tdiffs.dat | cut -f 3 -d " "`
sum=`echo "($tds - $tdr) ^ 2" | bc -l`
Sum=`echo "$sum + $Sum" | bc -l`
done
echo $Sum
}


The problem is the "Sum" variable does not accept its new value after the while loop ends.



 RMS_Cal
++ cat synt_tdiffs.dat
++ wc -l
+ num=1
+ local Sum=0
+ cat synt_tdiffs.dat
+ read info
++ echo AHWZ ILBA 25.631
++ cut -f 2 -d ' '
+ sta=ILBA
++ echo AHWZ ILBA 25.631
++ cut -f 3 -d ' '
+ tds=25.631
++ grep -w ILBA real_tdiffs.dat
++ cut -f 3 -d ' '
+ tdr=527
++ echo '(25.631 - 527) ^ 2'
++ bc -l
+ sum=251370.874161
++ echo '251370.874161 + 0'
++ bc -l
+ Sum=251370.874161
+ read info
+ echo 0
0


How can I handle this problem?









share














I wrote the following function to sequentially add sum calculation into a variable.



RMS_Cal ()
{
num=`cat synt_tdiffs.dat | wc -l`
Sum=0
cat synt_tdiffs.dat | while read info; do
sta=`echo $info | cut -f 2 -d " "`
tds=`echo $info | cut -f 3 -d " "`
tdr=`grep -w "$sta" real_tdiffs.dat | cut -f 3 -d " "`
sum=`echo "($tds - $tdr) ^ 2" | bc -l`
Sum=`echo "$sum + $Sum" | bc -l`
done
echo $Sum
}


The problem is the "Sum" variable does not accept its new value after the while loop ends.



 RMS_Cal
++ cat synt_tdiffs.dat
++ wc -l
+ num=1
+ local Sum=0
+ cat synt_tdiffs.dat
+ read info
++ echo AHWZ ILBA 25.631
++ cut -f 2 -d ' '
+ sta=ILBA
++ echo AHWZ ILBA 25.631
++ cut -f 3 -d ' '
+ tds=25.631
++ grep -w ILBA real_tdiffs.dat
++ cut -f 3 -d ' '
+ tdr=527
++ echo '(25.631 - 527) ^ 2'
++ bc -l
+ sum=251370.874161
++ echo '251370.874161 + 0'
++ bc -l
+ Sum=251370.874161
+ read info
+ echo 0
0


How can I handle this problem?







variable





share












share










share



share










asked 5 mins ago









alireza niksejelalireza niksejel

83 bronze badges




83 bronze badges
















  • Possible duplicate of Why is my variable local in one 'while read' loop, but not in another seemingly similar loop?

    – muru
    3 mins ago



















  • Possible duplicate of Why is my variable local in one 'while read' loop, but not in another seemingly similar loop?

    – muru
    3 mins ago

















Possible duplicate of Why is my variable local in one 'while read' loop, but not in another seemingly similar loop?

– muru
3 mins ago





Possible duplicate of Why is my variable local in one 'while read' loop, but not in another seemingly similar loop?

– muru
3 mins ago










0






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