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With the code int foobar(int*n){*n=*n+1; return*n;} when called int k=6; printf("k=%d\n",k)
what get printed? *
foobar (k)=6,*foobar (k)=7,k=*foobar (k)=6, k=7​

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With the code int foobar(int*n){*n=*n+1; return*n;} when called int k=6; printf("k=%d\n",k)
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