質問編集履歴
1
補足
title
CHANGED
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@@ -1,1 +1,1 @@
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1
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-
構造体
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1
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+
２つの構造体配列を比較しソート
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body
CHANGED
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@@ -15,12 +15,12 @@
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15
15
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typedef struct person09{
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char name[NAME_MAX];
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signed int profit;
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-
}
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+
}person09_t;
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typedef struct person10{
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char name[NAME_MAX];
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signed int profit;
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-
}
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}person10_t;
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void sort(int no, char *result[]){
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int i;
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@@ -45,53 +45,27 @@
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int no;
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int flag[5] = {0};
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char *result_array[8];
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48
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+
//char profit_array[8] = {0};
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char *area;
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int profit_d[8] = {0};
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52
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+
//構造体配列
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53
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person09_t ary_09[6] = {{"ｱﾄｳ",270},{"ｶｽｶﾞ",325},{"ﾀｹﾓﾄ",412},{"ﾀﾅｶ",333},{"ﾔﾏｼﾀ",277},{"ﾖｼｵｶ",192}};
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54
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+
person10_t ary_10[5] = {{"ｶｽｶﾞ",357},{"ｽｽﾞｷ",210},{"ﾀｹﾓﾄ",445},{"ﾔﾏｼﾀ",233},{"ﾖｼﾀﾞ",145}};
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55
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52
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-
//構造体宣言
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53
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-
person09 data09[6];
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54
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-
person10 data10[5];
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56
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57
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-
//構造体初期化
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58
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-
data09[0].profit = 270;
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59
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-
data09[1].profit = 325;
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60
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-
data09[2].profit = 412;
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-
data09[3].profit = 333;
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62
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-
data09[4].profit = 277;
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63
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-
data09[5].profit = 192;
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64
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-
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65
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-
strcpy(data09[0].name, "ｱﾄｳ");
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66
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-
strcpy(data09[1].name, "ｶｽｶﾞ");
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67
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-
strcpy(data09[2].name, "ﾀｹﾓﾄ");
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68
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-
strcpy(data09[3].name, "ﾀﾅｶ");
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69
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-
strcpy(data09[4].name, "ﾔﾏｼﾀ");
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70
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-
strcpy(data09[5].name, "ﾖｼｵｶ");
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71
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-
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-
data10[0].profit = 357;
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73
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-
data10[1].profit = 210;
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-
data10[2].profit = 445;
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75
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-
data10[3].profit = 233;
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76
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-
data10[4].profit = 145;
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-
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-
strcpy(data10[0].name, "ｶｽｶﾞ");
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-
strcpy(data10[1].name, "ｽｽﾞｷ");
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-
strcpy(data10[2].name, "ﾀｹﾓﾄ");
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-
strcpy(data10[3].name, "ﾔﾏｼﾀ");
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82
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-
strcpy(data10[4].name, "ﾖｼﾀﾞ");
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83
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-
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84
58
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//元の数字を表示
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85
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for( i=0 ; i < 6; i++){
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86
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-
printf("%s ",
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printf("%s ",ary_09[i].name);
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87
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-
printf("%d\n",
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61
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printf("%d\n",ary_09[i].profit);
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88
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}
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89
63
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90
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91
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printf("\n");
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92
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for( i=0 ; i < 5; i++){
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93
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-
printf("%s ",
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67
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printf("%s ",ary_10[i].name);
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94
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printf("%d\n",
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68
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printf("%d\n",ary_10[i].profit);
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95
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}
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70
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97
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printf("\n");
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@@ -106,12 +80,12 @@
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106
80
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for( j = 0; j < 5; j++){
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107
81
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108
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//売上比較
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109
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-
if(strcmp(
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83
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if(strcmp( ary_09[i].name, ary_10[j].name) == 0 ){
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110
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-
printf("%s ",
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84
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printf("%s ",ary_09[i].name);
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111
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-
profit_d[i] =
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85
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profit_d[i] = (ary_10[j].profit) - (ary_09[i].profit);
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112
86
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printf("%d\n",profit_d[i] );
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113
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flag[j] = CHECKED;
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114
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-
result_array[r] =
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88
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result_array[r] = ary_09[i].name;
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115
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r++;
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116
90
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break;
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117
91
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}else{
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@@ -120,8 +94,8 @@
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120
94
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121
95
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//退社判定
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122
96
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if( cnt == RETIRE ){
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123
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-
printf("%s 退社\n",
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97
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printf("%s 退社\n",ary_09[i].name);
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124
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-
result_array[r] =
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98
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result_array[r] = ary_09[i].name;
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125
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r++;
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126
100
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break;
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127
101
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}
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@@ -131,13 +105,12 @@
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131
105
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//入社判定
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132
106
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for( j = 0; j < 5; j++){
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133
107
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if( flag[j] != CHECKED ){
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134
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-
printf("%s 入社\n",
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108
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printf("%s 入社\n",ary_10[j].name);
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135
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-
result_array[r] =
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109
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result_array[r] = ary_10[j].name;
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136
110
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r++;
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137
111
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}
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138
112
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}
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139
113
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140
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-
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141
114
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//並び替え処理
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142
115
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printf("\n");
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143
116
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puts("---------------------------------------------");
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@@ -151,10 +124,13 @@
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151
124
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free(area);
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152
125
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return 0;
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153
126
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}
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127
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+
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128
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+
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129
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+
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130
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+
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154
131
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```
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155
132
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156
133
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昨年と今年の売り上げを比較し、昨年いない人は「入社」、今年いない人は「退社」というプログラムを書いています。
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157
134
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158
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-
名前を５０音順に並び替え、売り上げ、入社、退社
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135
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+
２つの構造体配列を比較し名前を５０音順に並び替え、売り上げ、入社、退社 を名前と紐づけたいです。
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159
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-
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160
136
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何か良い方法があったら教えて頂きたいです。
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