Vertical concatenation in Software

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582 Vertical concatenation
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What makes character pictures a fun example is that, once we have them, we can do things with them We just saw one operation framing a picture Another operation is concatenation, which we can do both vertically and horizontally We'll look at vertical concatenation here, and at horizontal concatenation in the next section Pictures are naturally organized by rows, in the sense that we represent a picture by a vector<string>, each element of which is a row Therefore, concatenating two pictures vertically is simple: We merely concatenate the vectors that represent them Doing so will cause the two pictures to line up along their left margins, which is a reasonable way to define vertical concatenation The only problem is that although there is a string concatenation operation, there is no vector concatenation operation As a result, we have to do the work ourselves:
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vector<string> vcat(const vector<string>& top, const vector<string>& bottom)
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This function uses only facilities that we have already seen: We define ret as a copy of top, append each element of bottom to ret, and return ret as its result The loop in this function implements one form of a common idea, namely, that of inserting a copy of elements from one container into another In this particular case, we are appending the elements, which we can think of as inserting them at the end Because this operation is so common, the library offers a way of doing it without writing a loop Instead of
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for (vector<string>::const_iterator it = bottombegin(); it != bottomend(); ++it) retpush_back(*it);
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retinsert(retend(), bottombegin(), bottomend());
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583 Horizontal concatenation
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By horizontal concatenation, we mean taking two pictures, and making a new picture in which one of the input pictures forms the left part of the new picture, and the other forms the right part Before we start, we need to think about what we want to do when the pictures to concatenate are different sizes We'll arbitrarily decide that we'll align them along their top edges Thus, each row of the output picture will be the result of concatenating the corresponding rows of the two input pictures We'll have to pad the left-hand picture's rows to make them take up the right amount of space in the output picture In addition to padding the left-hand picture, we also have to worry about what to do when the pictures have a different number of rows For example, if p holds our initial picture, we might want to concatenate the original value of p horizontally with the result of framing p That is, we'd like hcat(p, frame(p)) to produce
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Note that the left-hand picture has fewer rows than the right-hand picture This fact implies that we will have to pad the output on the left-hand side to account for these missing rows If the left-hand picture is longer, we'll just copy the strings from it into the new picture; we won't bother to pad the (empty) right side with blanks With this analysis complete, we can write our function:
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vector<string> hcat(const vector<string>& left, const vector<string>& right) { vector<string> ret; // add 1 to leave a space between pictures string::size_type width1 = width(left) + 1; // indices to look at elements from left and right respectively vector<string>::size_type i = 0, j = 0; // continue until we've seen all rows from both pictures while (i != leftsize() || j != rightsize()) { // construct new string to hold characters from both pictures string s;
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// copy a row from the left-hand side, if there is one if (i != leftsize()) s = left[i++]; // pad to full width s += string(width1 - ssize(), ' '); // copy a row from the right-hand side, if there is one if (j != rightsize()) s += right[j++]; // add s to the picture we're creating retpush_back(s); } return } ret;
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We start, as we did for frame and vcat, by defining the picture that we'll return Our next step is to compute the width to which we must pad the left-hand picture That width will be one more than the width of the picture itself, to leave a space between the pictures when we concatenate them Next, we iterate through both pictures, copying an element from the first, padded as necessary, followed by an element from the second The only tricky part is taking care of what to do if we run out of elements in one picture before we run out of elements in the other Our iteration continues until we have copied all the elements for each input vector Hence, the while loop continues until both indices reach the end of their respective pictures If we have not yet exhausted left, we copy its current element into s Regardless of whether we copied anything from left, we next call the string compound assignment operator, +=, to pad the output to the appropriate width The compound assignment operator defined by the string library operates as you might expect: It adds the right-hand operand to its left-hand operand and stores the result in the left-hand side Of course, "add" here means string concatenation We determine how much to pad by subtracting ssize() from width1 We know that either ssize() is the size of the string that we copied from left, or it is zero because there was no entry to copy In the first case, ssize() will be greater than zero and less than width1, because we added one to the length of the longest string to account for a space between the two pictures Thus, in this case, we'll append one or more blanks to s If ssize() is zero, then we'll pad the entire output line Having copied and padded the string for the left-hand picture, we need only append the string from the right-hand picture, assuming that there still is an element from right to copy Regardless of whether we added a value from right, we push s onto the output vector, and continue until we've processed both input vectors remembering to return to
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