This question's explanation and your answers so far.
関数binaryToIntegerは，1桁以上の符号なし2進数を文字列で表した値を引数binaryStrで受け取り，その値を整数に変換した結果を戻り値とする。例えば，引数として“100”を受け取ると，4を返す。プログラム中のa，bに入れる字句の適切な組合せはどれか。 〔プログラム〕 ○整数型: binaryToInteger(文字列型: binaryStr) 整数型: integerNum, digitNum, exponent, i integerNum ← 0 for (i を 1 から binaryStr の文字数 まで 1 ずつ増やす) digitNum ← binaryStr の末尾から i 番目の文字を整数型に変換した値 // 例: 文字“1”であれば整数値1に変換 exponent ← ［ a ］ integerNum ← ［ b ］ endfor return integerNum
a's “(2 to the power of i) − 1” is wrong. The weight of the i-th digit from the end is 2 to the power of (i − 1), not (2 to the power of i) − 1 (e.g., at i=1 the weight is 1, and this expression happens to give 2−1=1 which matches, but at i=2 it gives 3 instead of the correct 2). Furthermore, b's “integerNum × digitNum × exponent” is also wrong: multiplying the digits together always yields 0 because the initial value is 0, so the conversion cannot be done. Both a and b are inappropriate.
b's “integerNum + digitNum × exponent” is correct cumulative addition, but a's “(2 to the power of i) − 1” is wrong. The weight of the i-th digit from the end is 2 to the power of (i − 1) (1 at i=1, 2 at i=2, 4 at i=3), whereas (2 to the power of i) − 1 (3 at i=2, 7 at i=3) does not match the digit weights, so even substituting “100” does not yield 4.
a's “2 to the power of (i − 1)” is the correct digit weight, but b's “integerNum × digitNum × exponent” is wrong. Since integerNum starts at 0, updating it by multiplication keeps multiplying by 0 every time, making the return value always 0. The correct operation must be addition that adds in each digit's contribution.
正解 In binary, the weight of the rightmost digit (the 1st from the end) is 2 to the power of 0, the one to its left is 2 to the power of 1, and so on, increasing; therefore the weight of the i-th digit from the end is “2 to the power of (i − 1)”. This goes into a (exponent ← 2 to the power of (i − 1)). For each digit, the digit value digitNum multiplied by the weight exponent is added into the running total, so b becomes “integerNum + digitNum × exponent”, and the whole value is obtained by cumulative addition. Actually substituting “100”, at i=1 the weight gives 1×0=0, at i=2 the weight gives 2×0=0, at i=3 the weight gives 4×1=4, so integerNum is 4. This matches the question's return value 4, so combination エ is correct.