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我在while循環中有以下一段代碼,我用exp()函數計算了一些概率。無論在循環的第7次迭代中對程序 的輸入是什麼,exp都返回nan。exp()函數在它不應該返回的時候返回nan
if(new<=old){
coloring[random_node]=random_color;
}else{
proba=exp((-(new-old))/temperature);
/*assert(!isnan(proba));*/
printf("proba == %.50f\n",proba);
if(random_percent(proba)){
coloring[random_node]=random_color;
}
}
以下是循環內第6次和第7次迭代的調試信息。
Breakpoint 1, graph_coloring_local_search (objectiveValue=50, N=50, E=350, edge_list=0x804d170, node_list=0x804dc68, maxIterations=100,
initial_temperature=7) at coloring.c:391
391 proba=exp((-(new-old))/temperature);
(gdb) p new
$21 = 1
(gdb) p old
$22 = 0
(gdb) p temperature
$23 = 6.9992999999999999
(gdb) p -(new-old)/temperature
$24 = -0.14287143000014288
(gdb) p ((double(*)())exp)(-(new-old)/temperature)
$25 = 0.8668655146301385
(gdb) c
Continuing.
proba == 0.86686551463013850060690401733154430985450744628906
Breakpoint 1, graph_coloring_local_search (objectiveValue=50, N=50, E=350, edge_list=0x804d170, node_list=0x804dc68, maxIterations=100,
initial_temperature=7) at coloring.c:391
391 proba=exp((-(new-old))/temperature);
(gdb) p new
$26 = 1
(gdb) p old
$27 = 0
(gdb) p temperature
$28 = 6.9992999999999999
(gdb) p -(new-old)/temperature
$29 = -0.14287143000014288
(gdb) p ((double(*)())exp)(-(new-old)/temperature)
$30 = -nan(0x8000000000000)
(gdb) c
Continuing.
proba == -nan
在這兩種情況下,使用的變量都具有完全相同的值。
如果在GDB提示符下輸入'p((double(*)(double))exp)( - (new-old)/ temperature)',事情會改變嗎?如果是這樣,你是否#include'? (我聞到一個隱含的聲明。) –
zwol
哦,如果這不能解決問題,請嘗試'valgrind'。這不是顯而易見的內存損壞,但它可能是微妙的內存損壞。 – zwol
沒有什麼改變,我包括數學庫 – rex123