We have as unsigned character like obff and I want program to count no of 1s occur in that character.

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Prashanth Kannadaguli

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unsigned int v; // count the number of bits set in v unsigned int c; // c accumulates the total bits set in v for (c = 0; v; v >>= 1) { c += v & 1; } The naive approach requires one iteration per bit, until no more bits are set. So on a 32-bit word with only the high set, it will go through...
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unsigned int v; // count the number of bits set in v unsigned int c; // c accumulates the total bits set in v for (c = 0; v; v >>= 1) { c += v & 1; } The naive approach requires one iteration per bit, until no more bits are set. So on a 32-bit word with only the high set, it will go through 32 iterations. Let me know if you need more detailed explaination. read less
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You can make a loop attached with a counter and shift right each time. loop testing condition should be (obff!=0). and with the help of count variable you can count number of one's.
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Dear Siddhant, I am sending a link that can help you to understand the link please open and try to learn, if is there any problem must reply me.. and don't confuse in unsigned int and unsigned char.. you can(put) unsigned char instead of unsigned int..
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Write a if condition that checks the every index of the string with 1. If the condition is true then increase the count variable value by one. Count variable must declared at the beginning of the program itself.
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Write a if condition that checks each and every index of string with 1 until u reach null termination . If the condition is true then increase the value of count variable by one. Count variable must be declared at the beginning of the program. At last print the count variable. That's it. Have fun with...
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Write a if condition that checks each and every index of string with 1 until u reach null termination . If the condition is true then increase the value of count variable by one. Count variable must be declared at the beginning of the program. At last print the count variable. That's it. Have fun with programs. read less
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