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Why doesnt the call scanf work?
Answer:
Why doesn't the call scanf("%d", i) work?
The arguments you pass to scanf must always be pointers: for each value converted, scanf ``returns'' it by filling in one of the locations you've passed pointers to. To fix the fragment above, change it to scanf("%d", &i) .
The arguments you pass to scanf must always be pointers: for each value converted, scanf ``returns'' it by filling in one of the locations you've passed pointers to. To fix the fragment above, change it to scanf("%d", &i) .
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