programming languages Online Quiz - 273
Description: programming languages Online Quiz - 273 | |
Number of Questions: 20 | |
Created by: Aliensbrain Bot | |
Tags: programming languages |
Which of the following is not a task in Workflow Manager
Which of the following is/are an Active transformation
What types of data source(s) can be read with the source qualifier transformation?
Which activities below can be performed using Designer
Choose the correct option - Transformations which have only Input/Output ports
When would you use Joiner transformation
How many sources can be joined using Joiner transformation
Which are the different Lookup cache types
Which transformation is used to read data from Cobol files?
What are the types of lookup
What are the views available in the Workflow Monitor
Which of the following are Informatica PowerCenter Server Components
Which datatype cannot be converted to any other datatype
Which one of the transformations below does not use index and data cache
Repository Manager is used to
1 options linesize=80; 2 title 'Example 2. Test of proc freq'; 3 4 proc format; 5 value $fsex '1'='Male' 6 '2'='Female'; NOTE: Format $FSEX has been output. 7 run; NOTE: PROCEDURE FORMAT used: real time 0.28 seconds cpu time 0.04 seconds 8 9 data sample1; 10 infile 'example2.dat'; 11 input caseid 1-4 mathsat 6-8 score 10-11 sex $ 13 teach 15 12 format sex $fsex.; ______ ______ ______ 228 228 228 13 label caseid='Subject Number' 14 mathsat='SAT Math Score' 15 score='Final Calculus Score' 16 sex='Sex of Participant' 17 teach='Student Rated Teaching Ability'; 18 ERROR 228-185: Informat $FSEX is unknown. ERROR 228-185: Informat $FSEX is unknown. ERROR 228-185: Informat $FSEX is unknown. NOTE: The SAS System stopped processing this step because of errors. NOTE: SAS set option OBS=0 and will continue to check statements. This may cause NOTE: No observations in data set. WARNING: The data set WORK.SAMPLE1 may be incomplete. When this step was stopped there were 0 observations and 6 variables. What is the problem?
void main() { int i = 10; int *p=&i; p++; printf("%d",(p-&i)); }
Which of the following are valid statements in IDEAL?
void main() { int i = 10; long *p=&i; p++; printf("%d",(p-&i)); }
void main() { int i = 10; long *p=&i; p++; printf("%d",abs(&i-p)); }