anonymous
  • anonymous
Can some one debug it pls.. #include int main() { int one,two,ans=0; char sym; printf("Enter a number: "); scanf("%d/n",&"%d/n",&one); printf("Enter another number :"); scanf("%d/n",&two); printf("Enter an operator(+,-,*,/): "); scanf("%ch/n",&sym); if(sym=='+') ans=one+two; else if(sym=='-') ans=one-two; else if(sym=='*') ans=one*two; else ans=one/two; printf("The result is: ",ans); getch(); return 0; } It always show same result i.e -7188
Computer Science
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
scanf("%d\n", &"%d\n", $one); is this a typo?
anonymous
  • anonymous
also scanf("%ch\n", &sym); %c is the correct conversion specifier and not %ch :-(
anonymous
  • anonymous
instead of that line, you can use sym = getchar()

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anonymous
  • anonymous
Finally, you've got printf("The result is: ", ans) with no %d conversion specifier in the format string.
anonymous
  • anonymous
C is a lot complicated than C++. :P
anonymous
  • anonymous
not really. It's just that scanf() is so inconvenient
anonymous
  • anonymous
now it shows 4+4=1
anonymous
  • anonymous
http://ideone.com/bTFbT
anonymous
  • anonymous
hahaha I have a bug in there: one / 2 instead of one / two :-P
anonymous
  • anonymous
http://ideone.com/xCtvs

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